Book 2. Olympiad Geometry Methods

If points \(B\) and \(D\) lie on the same side of chord \(AC\), then \(\angle ABC\) and \(\angle ADC\) are equal as inscribed angles standing on the same chord. If they lie on opposite sides, the angles are supplementary. In both cases the oriented notation gives \(\angle ABC\equiv\angle ADC\pmod{180^\circ}\).

Draw the circle through \(A,B,C\). Every point of this circle, different from \(A\) and \(C\), sees chord \(AC\) under an angle equal to \(\angle ABC\) in oriented notation. By the condition, point \(D\) sees the same segment \(AC\) under the same oriented angle. Therefore \(D\) belongs to this circle, and points \(A,B,C,D\) are cyclic.

Let \(O\) be the centre of the circle. Radius \(OA\) is perpendicular to tangent \(t\). The central angle \(\angle AOB\) is twice the inscribed angle \(\angle ACB\). From \(OA\perp t\), the angle between \(t\) and \(AB\) equals half of the central angle, that is \(\angle ACB\). Hence \(\angle(t,AB)\equiv\angle ACB\pmod{180^\circ}\).

Angle \(\angle BAD\) stands on the arc \(BD\) not containing \(A\), while \(\angle BCD\) stands on the other arc \(BD\). These two arcs together form the whole circle, so the corresponding inscribed angles sum to \(180^\circ\).

Since \(B,C,D,E\) are cyclic, \(\angle BDE\equiv\angle BCE\pmod{180^\circ}\). But \(BD\) lies on line \(BA\), and \(CE\) lies on line \(CA\), so \(\angle ADE\equiv\angle ACB\pmod{180^\circ}\). Similarly, from \(\angle BED\equiv\angle BCD\), we get \(\angle AED\equiv\angle ABC\pmod{180^\circ}\).

Let \(t\) be the tangent to the circle at \(A\). By the tangent-chord theorem, \(\angle(t,AB)\equiv\angle ACB\pmod{180^\circ}\). By the condition, \(\angle(l,AB)\) has the same value. Therefore lines \(l\) and \(t\) coincide, and \(l\) is tangent.

Let \(t_1\) and \(t_2\) be the tangents to \(\omega_1\) and \(\omega_2\) at \(A\). By the tangent-chord theorem, \(\angle(t_1,AB)\equiv\angle ACB\), and \(\angle(t_2,AB)\equiv\angle ADB\). Thus \(\angle(t_1,t_2)\equiv\angle(t_1,AB)-\angle(t_2,AB)\equiv\angle ACB-\angle ADB\pmod{180^\circ}\).

Since \(BD\perp AC\), we have \(\angle BDC=90^\circ\). Since \(CE\perp AB\), we have \(\angle BEC=90^\circ\). Therefore \(B,C,D,E\) lie on the circle with diameter \(BC\). Now apply the antiparallel result inside the angle: from cyclicity of \(B,C,D,E\), it follows that \(\angle ADE\equiv\angle ACB\) and \(\angle AED\equiv\angle ABC\pmod{180^\circ}\).

In triangle \(BCP\), we have \(\angle PBC=\angle DBC=\angle CBD\) and \(\angle BCP=\angle BCA=\angle ACB\). Therefore \(\angle BPC=180^\circ-\angle CBD-\angle ACB\). Angles \(\angle APB\) and \(\angle BPC\) are supplementary, hence \(\angle APB=\angle ACB+\angle CBD\).

Angle \(\angle APB\) is the oriented angle between lines \(AD\) and \(BC\). Rotate from \(AD\) to \(AB\), and then from \(AB\) to \(BC\). We get \(\angle APB\equiv\angle(AD,AB)+\angle(AB,BC)\). But \(\angle(AD,AB)=\angle DAB\), and \(\angle(AB,BC)\equiv\angle ABC\pmod{180^\circ}\). Therefore \(\angle APB\equiv\angle DAB+\angle ABC\pmod{180^\circ}\).

Let \(O\) be the centre of the circle. Then \(OB\perp TB\) and \(OC\perp TC\). Since \(\angle BAC<90^\circ\), the smaller central angle \(\angle BOC\) standing on arc \(BC\) equals \(2\angle BAC\). The smaller angle between the tangents is supplementary to this central angle. Therefore it equals \(180^\circ-2\angle BAC\).

Since \(BH\perp AC\), the angle between \(BH\) and \(AB\) equals \(90^\circ-\angle A\). Since \(CH\perp AB\), the angle between \(CH\) and \(AC\) is also expressed through \(90^\circ\). Summing around the vertex gives \(\angle BHC=180^\circ-\angle BAC\). In other words, the angle between two altitudes is supplementary to the angle between the corresponding sides.

From cyclicity of \(B,C,D,E\), we get \(\angle ADE\equiv\angle ACB\) and \(\angle AED\equiv\angle ABC\). Therefore \(\triangle ADE\sim\triangle ACB\). The corresponding sides give \(\frac{AD}{AC}=\frac{AE}{AB}\). Cross-multiplying, we obtain \(AD\cdot AB=AE\cdot AC\).

Let \(t\) be the common tangent at \(A\). For the first circle, \(\angle(t,AB)\equiv\angle ACB\). For the second circle, since \(AD\) lies on the same line as \(AB\), we have \(\angle(t,AD)\equiv\angle AED\). Hence \(\angle ACB\equiv\angle AED\). Since \(A,C,E\) are collinear, this means that lines \(BC\) and \(DE\) form equal angles with line \(CE\). Therefore \(BC\parallel DE\).

Since \(A,B,C,E\) are cyclic, \(\angle(CE,EB)\equiv\angle(CA,AB)\). Since \(A,B,D,F\) are cyclic, \(\angle(DF,FB)\equiv\angle(DA,AB)\). But \(C,A,D\) are collinear, so \(\angle(CA,AB)\equiv\angle(DA,AB)\). Also \(E,B,F\) are collinear. Therefore \(CE\) and \(DF\) form equal angles with the same line, so \(CE\parallel DF\).

Let \(t_1\) be the tangent to \((ABD)\) at \(D\), and \(t_2\) the tangent to \((ACD)\) at \(D\). By the tangent-chord theorem, \(\angle(t_1,DB)\equiv\angle DAB\), and \(\angle(DC,t_2)\equiv\angle DAC\). Since \(DB\) and \(DC\) lie on one line, the angle between \(t_1\) and \(t_2\) equals \(\angle DAB+\angle DAC=\angle BAC\) in oriented notation.

Since \(AD\parallel BC\), angles \(\angle DAB\) and \(\angle ABC\) sum to \(180^\circ\). Since \(ABCD\) is cyclic, \(\angle DAB+\angle DCB=180^\circ\). Therefore \(\angle ABC=\angle DCB\). In a trapezoid, equality of the angles at one base implies equality of the legs: for example, drop the altitudes from \(A\) and \(D\) to \(BC\) and obtain two congruent right triangles. Hence \(AB=CD\).

By the tangent-chord theorem, \(\angle TAB\equiv\angle ACB\). Since \(T,C,B\) are collinear, \(\angle ACB=\angle TCA\). Hence \(\angle TAB=\angle TCA\). Similarly, \(\angle TAC\equiv\angle ABC=\angle TBA\). Therefore \(\triangle TAB\sim\triangle TCA\).

Let \(t_1,t_2\) be the tangents to the circles at \(A\). By the tangent-chord theorem, \(\angle(t_1,AB)\equiv\angle ACB\), and \(\angle(t_2,AB)\equiv\angle ADB\). Therefore the angle between the circles equals \(\angle ACB-\angle ADB\) in oriented notation. By the condition, \(t_1\perp t_2\), so this angle is \(90^\circ\) modulo \(180^\circ\). Hence \(\angle ACB-\angle ADB\equiv90^\circ\pmod{180^\circ}\).

By the tangent-chord theorem, the angle between the tangent at \(D\) and chord \(DB\) equals \(\angle DAB\). Since the tangent is parallel to \(AC\), and \(DB\) lies on \(BC\), the same angle equals \(\angle ACB\). Also \(D\) lies on \(BC\), so \(\angle DAB=\angle BAC\). Therefore \(\angle BAC=\angle ACB\).

Since \(A,B,C,E\) are cyclic, \(\angle(CE,EB)\equiv\angle(CA,AB)\). Since \(A,B,D,F\) are cyclic, \(\angle(DF,FB)\equiv\angle(DA,AB)\). But \(C,A,D\) are collinear, so the right-hand sides are equal. Therefore \(\angle(CE,EB)\equiv\angle(DF,FB)\). Since \(CE\parallel DF\), we get \(\angle(DF,EB)\equiv\angle(DF,FB)\), hence \(EB\parallel FB\). Both lines pass through \(B\), so \(E,B,F\) are collinear.

Let \(t_1\) and \(t_2\) be the tangents to circles \((ABC)\) and \((ADE)\) at \(A\). By the tangent-chord theorem, \(\angle(t_1,AB)\equiv\angle ACB\), and \(\angle(t_2,AD)\equiv\angle AED\). Since \(AB\) and \(AD\) are one line, the tangents coincide if and only if \(\angle ACB\equiv\angle AED\). But points \(A,C,E\) are collinear, so this angle equality is equivalent to \(BC\parallel DE\). Therefore the circles are tangent at \(A\) if and only if \(BC\parallel DE\).

The angle between two circles at \(D\) is the angle between their tangents at that point. By the angle fact already proved for \(D\in BC\), the angle between circles \((ABD)\) and \((ACD)\) equals \(\angle BAC\) in oriented notation. By the condition, the tangents are perpendicular, so this angle is \(90^\circ\). Therefore \(\angle BAC=90^\circ\).

For circle \((ABD)\), the angle between the tangent at \(D\) and chord \(DB\) equals \(\angle DAB=\angle BAC\). But this tangent is parallel to \(AC\), and \(DB\) lies on \(BC\), so the same angle equals \(\angle ACB\). Hence \(\angle A=\angle C\).

For circle \((ACD)\), the angle between the tangent at \(D\) and chord \(DC\) equals \(\angle DAC=\angle BAC\). This tangent is parallel to \(AB\), and \(DC\) lies on \(BC\), so the same angle equals \(\angle ABC\). Hence \(\angle A=\angle B\). Thus \(\angle A=\angle B=\angle C\), and triangle \(ABC\) is equilateral.

E. Full solution. Since \(B,D,P,M\) lie on one circle, \(\angle BMP=\angle BDP\): both angles subtend chord \(BP\).

Points \(B,D,A\) are collinear, and points \(D,P,C\) are collinear. Hence \(\angle BDP\) is the oriented angle between lines \(AB\) and \(CD\).

Similarly, from the cyclic quadrilateral \(C,E,P,M\), we get \(\angle PME=\angle PCE\). But \(P,C,D\) are collinear and \(C,E,A\) are collinear, so \(\angle PCE\) is the oriented angle between \(CD\) and \(AC\).

Adding these two angles, we see that \(\angle BME\) equals the angle between \(AB\) and \(AC\), that is \(\angle BAC=49^\circ\).

But \(\angle BAE=\angle BAC\), since \(E\in AC\). Therefore \(\angle BME=\angle BAE\). By the converse criterion for equal inscribed angles, \(A,B,E,M\) lie on one circle.

Method comment. the second intersection of the two circles is a Miquel point and must also lie on the circle through \(A,B,E\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(PX\perp BC\) and \(PY\perp CA\), angles \(\angle PXC\) and \(\angle PYC\) are right angles. Therefore \(P,X,C,Y\) lie on one circle.

Similarly, \(PZ\perp AB\) and \(PY\perp CA\), so \(\angle PZA=\angle PYA=90^\circ\), and points \(P,Z,A,Y\) also lie on one circle.

From the first circle, \(\angle XYC=\angle XPC\). From the second circle, \(\angle ZYA=\angle ZPA\).

Now compare the right-hand sides. Since \(PX\perp BC\), \(\angle XPC=90^\circ-\angle BCP\). Since \(PZ\perp AB\), \(\angle ZPA=90^\circ-\angle BAP\).

Points \(A,B,C,P\) lie on one circle, so \(\angle BCP=\angle BAP\): these are inscribed angles subtending chord \(BP\). Hence \(\angle XPC=\angle ZPA\), and therefore \(\angle XYC=\angle ZYA\).

Since \(A,Y,C\) are collinear, equality \(\angle XYC=\angle ZYA\) means that rays \(YX\) and \(YZ\) lie on one line. Thus \(X,Y,Z\) are collinear.

Method comment. two circles with diameters \(PC\) and \(PA\) give equal angles at point \(Y\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since the tangent segments are equal, \(PB=PC\). Hence triangle \(PBC\) is isosceles, so the median \(PM\) to base \(BC\) is an altitude: \(PM\perp BC\).

Because \(PE\perp AC\), we have \(\angle PEC=90^\circ\). Also \(\angle PMC=90^\circ\). Therefore \(M,C,E,P\) lie on one circle.

In this circle, \(\angle MEP=\angle MCP\), since both angles subtend chord \(MP\). But \(CP\) is tangent to \(\Omega\) at \(C\), so by the tangent-chord theorem \(\angle MCP=\angle BAC\).

Since \(PE\perp AC\), we get \(\angle MEA=90^\circ-\angle MEP=90^\circ-\angle BAC\). Also \(\angle DAE=\angle BAC\), because \(D\in AB\) and \(E\in AC\). Thus \(\angle MEA+\angle DAE=90^\circ\), so \(ME\perp AD\).

Similarly, using the cyclic quadrilateral \(M,D,B,P\) and the tangent \(PB\), we obtain \(MD\perp AE\). Thus in triangle \(ADE\), the lines \(ME\) and \(MD\) are altitudes. Their intersection is \(M\), so \(M\) is the orthocenter of triangle \(ADE\).

Method comment. quadrilaterals \(MCEP\) and \(MDBP\) are cyclic, and the tangent converts angles at \(C\) and \(B\) into angles of the original triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(B,D,P,M\) lie on one circle, \(\angle BMP=\angle BDP\): both angles subtend chord \(BP\).

Points \(B,D,A\) are collinear, and points \(D,P,C\) are collinear. Hence \(\angle BDP\) is the oriented angle between lines \(AB\) and \(CD\).

Similarly, from the cyclic quadrilateral \(C,E,P,M\), we get \(\angle PME=\angle PCE\). But \(P,C,D\) are collinear and \(C,E,A\) are collinear, so \(\angle PCE\) is the oriented angle between \(CD\) and \(AC\).

Adding these two angles, we see that \(\angle BME\) equals the angle between \(AB\) and \(AC\), that is \(\angle BAC=70^\circ\).

But \(\angle BAE=\angle BAC\), since \(E\in AC\). Therefore \(\angle BME=\angle BAE\). By the converse criterion for equal inscribed angles, \(A,B,E,M\) lie on one circle.

Method comment. the second intersection of the two circles is a Miquel point and must also lie on the circle through \(A,B,E\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(PX\perp BC\) and \(PY\perp CA\), angles \(\angle PXC\) and \(\angle PYC\) are right angles. Therefore \(P,X,C,Y\) lie on one circle.

Similarly, \(PZ\perp AB\) and \(PY\perp CA\), so \(\angle PZA=\angle PYA=90^\circ\), and points \(P,Z,A,Y\) also lie on one circle.

From the first circle, \(\angle XYC=\angle XPC\). From the second circle, \(\angle ZYA=\angle ZPA\).

Now compare the right-hand sides. Since \(PX\perp BC\), \(\angle XPC=90^\circ-\angle BCP\). Since \(PZ\perp AB\), \(\angle ZPA=90^\circ-\angle BAP\).

Points \(A,B,C,P\) lie on one circle, so \(\angle BCP=\angle BAP\): these are inscribed angles subtending chord \(BP\). Hence \(\angle XPC=\angle ZPA\), and therefore \(\angle XYC=\angle ZYA\).

Since \(A,Y,C\) are collinear, equality \(\angle XYC=\angle ZYA\) means that rays \(YX\) and \(YZ\) lie on one line. Thus \(X,Y,Z\) are collinear.

Method comment. two circles with diameters \(PC\) and \(PA\) give equal angles at point \(Y\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since the tangent segments are equal, \(PB=PC\). Hence triangle \(PBC\) is isosceles, so the median \(PM\) to base \(BC\) is an altitude: \(PM\perp BC\).

Because \(PE\perp AC\), we have \(\angle PEC=90^\circ\). Also \(\angle PMC=90^\circ\). Therefore \(M,C,E,P\) lie on one circle.

In this circle, \(\angle MEP=\angle MCP\), since both angles subtend chord \(MP\). But \(CP\) is tangent to \(\Omega\) at \(C\), so by the tangent-chord theorem \(\angle MCP=\angle BAC\).

Since \(PE\perp AC\), we get \(\angle MEA=90^\circ-\angle MEP=90^\circ-\angle BAC\). Also \(\angle DAE=\angle BAC\), because \(D\in AB\) and \(E\in AC\). Thus \(\angle MEA+\angle DAE=90^\circ\), so \(ME\perp AD\).

Similarly, using the cyclic quadrilateral \(M,D,B,P\) and the tangent \(PB\), we obtain \(MD\perp AE\). Thus in triangle \(ADE\), the lines \(ME\) and \(MD\) are altitudes. Their intersection is \(M\), so \(M\) is the orthocenter of triangle \(ADE\).

Method comment. quadrilaterals \(MCEP\) and \(MDBP\) are cyclic, and the tangent converts angles at \(C\) and \(B\) into angles of the original triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(B,D,P,M\) lie on one circle, \(\angle BMP=\angle BDP\): both angles subtend chord \(BP\).

Points \(B,D,A\) are collinear, and points \(D,P,C\) are collinear. Hence \(\angle BDP\) is the oriented angle between lines \(AB\) and \(CD\).

Similarly, from the cyclic quadrilateral \(C,E,P,M\), we get \(\angle PME=\angle PCE\). But \(P,C,D\) are collinear and \(C,E,A\) are collinear, so \(\angle PCE\) is the oriented angle between \(CD\) and \(AC\).

Adding these two angles, we see that \(\angle BME\) equals the angle between \(AB\) and \(AC\), that is \(\angle BAC=48^\circ\).

But \(\angle BAE=\angle BAC\), since \(E\in AC\). Therefore \(\angle BME=\angle BAE\). By the converse criterion for equal inscribed angles, \(A,B,E,M\) lie on one circle.

Method comment. the second intersection of the two circles is a Miquel point and must also lie on the circle through \(A,B,E\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(PX\perp BC\) and \(PY\perp CA\), angles \(\angle PXC\) and \(\angle PYC\) are right angles. Therefore \(P,X,C,Y\) lie on one circle.

Similarly, \(PZ\perp AB\) and \(PY\perp CA\), so \(\angle PZA=\angle PYA=90^\circ\), and points \(P,Z,A,Y\) also lie on one circle.

From the first circle, \(\angle XYC=\angle XPC\). From the second circle, \(\angle ZYA=\angle ZPA\).

Now compare the right-hand sides. Since \(PX\perp BC\), \(\angle XPC=90^\circ-\angle BCP\). Since \(PZ\perp AB\), \(\angle ZPA=90^\circ-\angle BAP\).

Points \(A,B,C,P\) lie on one circle, so \(\angle BCP=\angle BAP\): these are inscribed angles subtending chord \(BP\). Hence \(\angle XPC=\angle ZPA\), and therefore \(\angle XYC=\angle ZYA\).

Since \(A,Y,C\) are collinear, equality \(\angle XYC=\angle ZYA\) means that rays \(YX\) and \(YZ\) lie on one line. Thus \(X,Y,Z\) are collinear.

Method comment. two circles with diameters \(PC\) and \(PA\) give equal angles at point \(Y\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since the tangent segments are equal, \(PB=PC\). Hence triangle \(PBC\) is isosceles, so the median \(PM\) to base \(BC\) is an altitude: \(PM\perp BC\).

Because \(PE\perp AC\), we have \(\angle PEC=90^\circ\). Also \(\angle PMC=90^\circ\). Therefore \(M,C,E,P\) lie on one circle.

In this circle, \(\angle MEP=\angle MCP\), since both angles subtend chord \(MP\). But \(CP\) is tangent to \(\Omega\) at \(C\), so by the tangent-chord theorem \(\angle MCP=\angle BAC\).

Since \(PE\perp AC\), we get \(\angle MEA=90^\circ-\angle MEP=90^\circ-\angle BAC\). Also \(\angle DAE=\angle BAC\), because \(D\in AB\) and \(E\in AC\). Thus \(\angle MEA+\angle DAE=90^\circ\), so \(ME\perp AD\).

Similarly, using the cyclic quadrilateral \(M,D,B,P\) and the tangent \(PB\), we obtain \(MD\perp AE\). Thus in triangle \(ADE\), the lines \(ME\) and \(MD\) are altitudes. Their intersection is \(M\), so \(M\) is the orthocenter of triangle \(ADE\).

Method comment. quadrilaterals \(MCEP\) and \(MDBP\) are cyclic, and the tangent converts angles at \(C\) and \(B\) into angles of the original triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(B,D,P,M\) lie on one circle, \(\angle BMP=\angle BDP\): both angles subtend chord \(BP\).

Points \(B,D,A\) are collinear, and points \(D,P,C\) are collinear. Hence \(\angle BDP\) is the oriented angle between lines \(AB\) and \(CD\).

Similarly, from the cyclic quadrilateral \(C,E,P,M\), we get \(\angle PME=\angle PCE\). But \(P,C,D\) are collinear and \(C,E,A\) are collinear, so \(\angle PCE\) is the oriented angle between \(CD\) and \(AC\).

Adding these two angles, we see that \(\angle BME\) equals the angle between \(AB\) and \(AC\), that is \(\angle BAC=69^\circ\).

But \(\angle BAE=\angle BAC\), since \(E\in AC\). Therefore \(\angle BME=\angle BAE\). By the converse criterion for equal inscribed angles, \(A,B,E,M\) lie on one circle.

Method comment. the second intersection of the two circles is a Miquel point and must also lie on the circle through \(A,B,E\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(PX\perp BC\) and \(PY\perp CA\), angles \(\angle PXC\) and \(\angle PYC\) are right angles. Therefore \(P,X,C,Y\) lie on one circle.

Similarly, \(PZ\perp AB\) and \(PY\perp CA\), so \(\angle PZA=\angle PYA=90^\circ\), and points \(P,Z,A,Y\) also lie on one circle.

From the first circle, \(\angle XYC=\angle XPC\). From the second circle, \(\angle ZYA=\angle ZPA\).

Now compare the right-hand sides. Since \(PX\perp BC\), \(\angle XPC=90^\circ-\angle BCP\). Since \(PZ\perp AB\), \(\angle ZPA=90^\circ-\angle BAP\).

Points \(A,B,C,P\) lie on one circle, so \(\angle BCP=\angle BAP\): these are inscribed angles subtending chord \(BP\). Hence \(\angle XPC=\angle ZPA\), and therefore \(\angle XYC=\angle ZYA\).

Since \(A,Y,C\) are collinear, equality \(\angle XYC=\angle ZYA\) means that rays \(YX\) and \(YZ\) lie on one line. Thus \(X,Y,Z\) are collinear.

Method comment. two circles with diameters \(PC\) and \(PA\) give equal angles at point \(Y\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since the tangent segments are equal, \(PB=PC\). Hence triangle \(PBC\) is isosceles, so the median \(PM\) to base \(BC\) is an altitude: \(PM\perp BC\).

Because \(PE\perp AC\), we have \(\angle PEC=90^\circ\). Also \(\angle PMC=90^\circ\). Therefore \(M,C,E,P\) lie on one circle.

In this circle, \(\angle MEP=\angle MCP\), since both angles subtend chord \(MP\). But \(CP\) is tangent to \(\Omega\) at \(C\), so by the tangent-chord theorem \(\angle MCP=\angle BAC\).

Since \(PE\perp AC\), we get \(\angle MEA=90^\circ-\angle MEP=90^\circ-\angle BAC\). Also \(\angle DAE=\angle BAC\), because \(D\in AB\) and \(E\in AC\). Thus \(\angle MEA+\angle DAE=90^\circ\), so \(ME\perp AD\).

Similarly, using the cyclic quadrilateral \(M,D,B,P\) and the tangent \(PB\), we obtain \(MD\perp AE\). Thus in triangle \(ADE\), the lines \(ME\) and \(MD\) are altitudes. Their intersection is \(M\), so \(M\) is the orthocenter of triangle \(ADE\).

Method comment. quadrilaterals \(MCEP\) and \(MDBP\) are cyclic, and the tangent converts angles at \(C\) and \(B\) into angles of the original triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since the tangent segments are equal, \(PB=PC\). Hence triangle \(PBC\) is isosceles, so the median \(PM\) to base \(BC\) is an altitude: \(PM\perp BC\).

Because \(PE\perp AC\), we have \(\angle PEC=90^\circ\). Also \(\angle PMC=90^\circ\). Therefore \(M,C,E,P\) lie on one circle.

In this circle, \(\angle MEP=\angle MCP\), since both angles subtend chord \(MP\). But \(CP\) is tangent to \(\Omega\) at \(C\), so by the tangent-chord theorem \(\angle MCP=\angle BAC\).

Since \(PE\perp AC\), we get \(\angle MEA=90^\circ-\angle MEP=90^\circ-\angle BAC\). Also \(\angle DAE=\angle BAC\), because \(D\in AB\) and \(E\in AC\). Thus \(\angle MEA+\angle DAE=90^\circ\), so \(ME\perp AD\).

Similarly, using the cyclic quadrilateral \(M,D,B,P\) and the tangent \(PB\), we obtain \(MD\perp AE\). Thus in triangle \(ADE\), the lines \(ME\) and \(MD\) are altitudes. Their intersection is \(M\), so \(M\) is the orthocenter of triangle \(ADE\).

Method comment. quadrilaterals \(MCEP\) and \(MDBP\) are cyclic, and the tangent converts angles at \(C\) and \(B\) into angles of the original triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(PX\perp BC\) and \(PY\perp CA\), angles \(\angle PXC\) and \(\angle PYC\) are right angles. Therefore \(P,X,C,Y\) lie on one circle.

Similarly, \(PZ\perp AB\) and \(PY\perp CA\), so \(\angle PZA=\angle PYA=90^\circ\), and points \(P,Z,A,Y\) also lie on one circle.

From the first circle, \(\angle XYC=\angle XPC\). From the second circle, \(\angle ZYA=\angle ZPA\).

Now compare the right-hand sides. Since \(PX\perp BC\), \(\angle XPC=90^\circ-\angle BCP\). Since \(PZ\perp AB\), \(\angle ZPA=90^\circ-\angle BAP\).

Points \(A,B,C,P\) lie on one circle, so \(\angle BCP=\angle BAP\): these are inscribed angles subtending chord \(BP\). Hence \(\angle XPC=\angle ZPA\), and therefore \(\angle XYC=\angle ZYA\).

Since \(A,Y,C\) are collinear, equality \(\angle XYC=\angle ZYA\) means that rays \(YX\) and \(YZ\) lie on one line. Thus \(X,Y,Z\) are collinear.

Method comment. two circles with diameters \(PC\) and \(PA\) give equal angles at point \(Y\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since the tangent segments are equal, \(PB=PC\). Hence triangle \(PBC\) is isosceles, so the median \(PM\) to base \(BC\) is an altitude: \(PM\perp BC\).

Because \(PE\perp AC\), we have \(\angle PEC=90^\circ\). Also \(\angle PMC=90^\circ\). Therefore \(M,C,E,P\) lie on one circle.

In this circle, \(\angle MEP=\angle MCP\), since both angles subtend chord \(MP\). But \(CP\) is tangent to \(\Omega\) at \(C\), so by the tangent-chord theorem \(\angle MCP=\angle BAC\).

Since \(PE\perp AC\), we get \(\angle MEA=90^\circ-\angle MEP=90^\circ-\angle BAC\). Also \(\angle DAE=\angle BAC\), because \(D\in AB\) and \(E\in AC\). Thus \(\angle MEA+\angle DAE=90^\circ\), so \(ME\perp AD\).

Similarly, using the cyclic quadrilateral \(M,D,B,P\) and the tangent \(PB\), we obtain \(MD\perp AE\). Thus in triangle \(ADE\), the lines \(ME\) and \(MD\) are altitudes. Their intersection is \(M\), so \(M\) is the orthocenter of triangle \(ADE\).

Method comment. quadrilaterals \(MCEP\) and \(MDBP\) are cyclic, and the tangent converts angles at \(C\) and \(B\) into angles of the original triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(B,D,P,M\) lie on one circle, \(\angle BMP=\angle BDP\): both angles subtend chord \(BP\).

Points \(B,D,A\) are collinear, and points \(D,P,C\) are collinear. Hence \(\angle BDP\) is the oriented angle between lines \(AB\) and \(CD\).

Similarly, from the cyclic quadrilateral \(C,E,P,M\), we get \(\angle PME=\angle PCE\). But \(P,C,D\) are collinear and \(C,E,A\) are collinear, so \(\angle PCE\) is the oriented angle between \(CD\) and \(AC\).

Adding these two angles, we see that \(\angle BME\) equals the angle between \(AB\) and \(AC\), that is \(\angle BAC=68^\circ\).

But \(\angle BAE=\angle BAC\), since \(E\in AC\). Therefore \(\angle BME=\angle BAE\). By the converse criterion for equal inscribed angles, \(A,B,E,M\) lie on one circle.

Method comment. the second intersection of the two circles is a Miquel point and must also lie on the circle through \(A,B,E\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(PX\perp BC\) and \(PY\perp CA\), angles \(\angle PXC\) and \(\angle PYC\) are right angles. Therefore \(P,X,C,Y\) lie on one circle.

Similarly, \(PZ\perp AB\) and \(PY\perp CA\), so \(\angle PZA=\angle PYA=90^\circ\), and points \(P,Z,A,Y\) also lie on one circle.

From the first circle, \(\angle XYC=\angle XPC\). From the second circle, \(\angle ZYA=\angle ZPA\).

Now compare the right-hand sides. Since \(PX\perp BC\), \(\angle XPC=90^\circ-\angle BCP\). Since \(PZ\perp AB\), \(\angle ZPA=90^\circ-\angle BAP\).

Points \(A,B,C,P\) lie on one circle, so \(\angle BCP=\angle BAP\): these are inscribed angles subtending chord \(BP\). Hence \(\angle XPC=\angle ZPA\), and therefore \(\angle XYC=\angle ZYA\).

Since \(A,Y,C\) are collinear, equality \(\angle XYC=\angle ZYA\) means that rays \(YX\) and \(YZ\) lie on one line. Thus \(X,Y,Z\) are collinear.

Method comment. two circles with diameters \(PC\) and \(PA\) give equal angles at point \(Y\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(B,D,P,M\) lie on one circle, \(\angle BMP=\angle BDP\): both angles subtend chord \(BP\).

Points \(B,D,A\) are collinear, and points \(D,P,C\) are collinear. Hence \(\angle BDP\) is the oriented angle between lines \(AB\) and \(CD\).

Similarly, from the cyclic quadrilateral \(C,E,P,M\), we get \(\angle PME=\angle PCE\). But \(P,C,D\) are collinear and \(C,E,A\) are collinear, so \(\angle PCE\) is the oriented angle between \(CD\) and \(AC\).

Adding these two angles, we see that \(\angle BME\) equals the angle between \(AB\) and \(AC\), that is \(\angle BAC=82^\circ\).

But \(\angle BAE=\angle BAC\), since \(E\in AC\). Therefore \(\angle BME=\angle BAE\). By the converse criterion for equal inscribed angles, \(A,B,E,M\) lie on one circle.

Method comment. the second intersection of the two circles is a Miquel point and must also lie on the circle through \(A,B,E\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since the tangent segments are equal, \(PB=PC\). Hence triangle \(PBC\) is isosceles, so the median \(PM\) to base \(BC\) is an altitude: \(PM\perp BC\).

Because \(PE\perp AC\), we have \(\angle PEC=90^\circ\). Also \(\angle PMC=90^\circ\). Therefore \(M,C,E,P\) lie on one circle.

In this circle, \(\angle MEP=\angle MCP\), since both angles subtend chord \(MP\). But \(CP\) is tangent to \(\Omega\) at \(C\), so by the tangent-chord theorem \(\angle MCP=\angle BAC\).

Since \(PE\perp AC\), we get \(\angle MEA=90^\circ-\angle MEP=90^\circ-\angle BAC\). Also \(\angle DAE=\angle BAC\), because \(D\in AB\) and \(E\in AC\). Thus \(\angle MEA+\angle DAE=90^\circ\), so \(ME\perp AD\).

Similarly, using the cyclic quadrilateral \(M,D,B,P\) and the tangent \(PB\), we obtain \(MD\perp AE\). Thus in triangle \(ADE\), the lines \(ME\) and \(MD\) are altitudes. Their intersection is \(M\), so \(M\) is the orthocenter of triangle \(ADE\).

Method comment. quadrilaterals \(MCEP\) and \(MDBP\) are cyclic, and the tangent converts angles at \(C\) and \(B\) into angles of the original triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(PX\perp BC\) and \(PY\perp CA\), angles \(\angle PXC\) and \(\angle PYC\) are right angles. Therefore \(P,X,C,Y\) lie on one circle.

Similarly, \(PZ\perp AB\) and \(PY\perp CA\), so \(\angle PZA=\angle PYA=90^\circ\), and points \(P,Z,A,Y\) also lie on one circle.

From the first circle, \(\angle XYC=\angle XPC\). From the second circle, \(\angle ZYA=\angle ZPA\).

Now compare the right-hand sides. Since \(PX\perp BC\), \(\angle XPC=90^\circ-\angle BCP\). Since \(PZ\perp AB\), \(\angle ZPA=90^\circ-\angle BAP\).

Points \(A,B,C,P\) lie on one circle, so \(\angle BCP=\angle BAP\): these are inscribed angles subtending chord \(BP\). Hence \(\angle XPC=\angle ZPA\), and therefore \(\angle XYC=\angle ZYA\).

Since \(A,Y,C\) are collinear, equality \(\angle XYC=\angle ZYA\) means that rays \(YX\) and \(YZ\) lie on one line. Thus \(X,Y,Z\) are collinear.

Method comment. two circles with diameters \(PC\) and \(PA\) give equal angles at point \(Y\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since the tangent segments are equal, \(PB=PC\). Hence triangle \(PBC\) is isosceles, so the median \(PM\) to base \(BC\) is an altitude: \(PM\perp BC\).

Because \(PE\perp AC\), we have \(\angle PEC=90^\circ\). Also \(\angle PMC=90^\circ\). Therefore \(M,C,E,P\) lie on one circle.

In this circle, \(\angle MEP=\angle MCP\), since both angles subtend chord \(MP\). But \(CP\) is tangent to \(\Omega\) at \(C\), so by the tangent-chord theorem \(\angle MCP=\angle BAC\).

Since \(PE\perp AC\), we get \(\angle MEA=90^\circ-\angle MEP=90^\circ-\angle BAC\). Also \(\angle DAE=\angle BAC\), because \(D\in AB\) and \(E\in AC\). Thus \(\angle MEA+\angle DAE=90^\circ\), so \(ME\perp AD\).

Similarly, using the cyclic quadrilateral \(M,D,B,P\) and the tangent \(PB\), we obtain \(MD\perp AE\). Thus in triangle \(ADE\), the lines \(ME\) and \(MD\) are altitudes. Their intersection is \(M\), so \(M\) is the orthocenter of triangle \(ADE\).

Method comment. quadrilaterals \(MCEP\) and \(MDBP\) are cyclic, and the tangent converts angles at \(C\) and \(B\) into angles of the original triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(B,D,P,M\) lie on one circle, \(\angle BMP=\angle BDP\): both angles subtend chord \(BP\).

Points \(B,D,A\) are collinear, and points \(D,P,C\) are collinear. Hence \(\angle BDP\) is the oriented angle between lines \(AB\) and \(CD\).

Similarly, from the cyclic quadrilateral \(C,E,P,M\), we get \(\angle PME=\angle PCE\). But \(P,C,D\) are collinear and \(C,E,A\) are collinear, so \(\angle PCE\) is the oriented angle between \(CD\) and \(AC\).

Adding these two angles, we see that \(\angle BME\) equals the angle between \(AB\) and \(AC\), that is \(\angle BAC=67^\circ\).

But \(\angle BAE=\angle BAC\), since \(E\in AC\). Therefore \(\angle BME=\angle BAE\). By the converse criterion for equal inscribed angles, \(A,B,E,M\) lie on one circle.

Method comment. the second intersection of the two circles is a Miquel point and must also lie on the circle through \(A,B,E\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(PX\perp BC\) and \(PY\perp CA\), angles \(\angle PXC\) and \(\angle PYC\) are right angles. Therefore \(P,X,C,Y\) lie on one circle.

Similarly, \(PZ\perp AB\) and \(PY\perp CA\), so \(\angle PZA=\angle PYA=90^\circ\), and points \(P,Z,A,Y\) also lie on one circle.

From the first circle, \(\angle XYC=\angle XPC\). From the second circle, \(\angle ZYA=\angle ZPA\).

Now compare the right-hand sides. Since \(PX\perp BC\), \(\angle XPC=90^\circ-\angle BCP\). Since \(PZ\perp AB\), \(\angle ZPA=90^\circ-\angle BAP\).

Points \(A,B,C,P\) lie on one circle, so \(\angle BCP=\angle BAP\): these are inscribed angles subtending chord \(BP\). Hence \(\angle XPC=\angle ZPA\), and therefore \(\angle XYC=\angle ZYA\).

Since \(A,Y,C\) are collinear, equality \(\angle XYC=\angle ZYA\) means that rays \(YX\) and \(YZ\) lie on one line. Thus \(X,Y,Z\) are collinear.

Method comment. two circles with diameters \(PC\) and \(PA\) give equal angles at point \(Y\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since the tangent segments are equal, \(PB=PC\). Hence triangle \(PBC\) is isosceles, so the median \(PM\) to base \(BC\) is an altitude: \(PM\perp BC\).

Because \(PE\perp AC\), we have \(\angle PEC=90^\circ\). Also \(\angle PMC=90^\circ\). Therefore \(M,C,E,P\) lie on one circle.

In this circle, \(\angle MEP=\angle MCP\), since both angles subtend chord \(MP\). But \(CP\) is tangent to \(\Omega\) at \(C\), so by the tangent-chord theorem \(\angle MCP=\angle BAC\).

Since \(PE\perp AC\), we get \(\angle MEA=90^\circ-\angle MEP=90^\circ-\angle BAC\). Also \(\angle DAE=\angle BAC\), because \(D\in AB\) and \(E\in AC\). Thus \(\angle MEA+\angle DAE=90^\circ\), so \(ME\perp AD\).

Similarly, using the cyclic quadrilateral \(M,D,B,P\) and the tangent \(PB\), we obtain \(MD\perp AE\). Thus in triangle \(ADE\), the lines \(ME\) and \(MD\) are altitudes. Their intersection is \(M\), so \(M\) is the orthocenter of triangle \(ADE\).

Method comment. quadrilaterals \(MCEP\) and \(MDBP\) are cyclic, and the tangent converts angles at \(C\) and \(B\) into angles of the original triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since the tangent segments are equal, \(PB=PC\). Hence triangle \(PBC\) is isosceles, so the median \(PM\) to base \(BC\) is an altitude: \(PM\perp BC\).

Because \(PE\perp AC\), we have \(\angle PEC=90^\circ\). Also \(\angle PMC=90^\circ\). Therefore \(M,C,E,P\) lie on one circle.

In this circle, \(\angle MEP=\angle MCP\), since both angles subtend chord \(MP\). But \(CP\) is tangent to \(\Omega\) at \(C\), so by the tangent-chord theorem \(\angle MCP=\angle BAC\).

Since \(PE\perp AC\), we get \(\angle MEA=90^\circ-\angle MEP=90^\circ-\angle BAC\). Also \(\angle DAE=\angle BAC\), because \(D\in AB\) and \(E\in AC\). Thus \(\angle MEA+\angle DAE=90^\circ\), so \(ME\perp AD\).

Similarly, using the cyclic quadrilateral \(M,D,B,P\) and the tangent \(PB\), we obtain \(MD\perp AE\). Thus in triangle \(ADE\), the lines \(ME\) and \(MD\) are altitudes. Their intersection is \(M\), so \(M\) is the orthocenter of triangle \(ADE\).

Method comment. quadrilaterals \(MCEP\) and \(MDBP\) are cyclic, and the tangent converts angles at \(C\) and \(B\) into angles of the original triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(PX\perp BC\) and \(PY\perp CA\), angles \(\angle PXC\) and \(\angle PYC\) are right angles. Therefore \(P,X,C,Y\) lie on one circle.

Similarly, \(PZ\perp AB\) and \(PY\perp CA\), so \(\angle PZA=\angle PYA=90^\circ\), and points \(P,Z,A,Y\) also lie on one circle.

From the first circle, \(\angle XYC=\angle XPC\). From the second circle, \(\angle ZYA=\angle ZPA\).

Now compare the right-hand sides. Since \(PX\perp BC\), \(\angle XPC=90^\circ-\angle BCP\). Since \(PZ\perp AB\), \(\angle ZPA=90^\circ-\angle BAP\).

Points \(A,B,C,P\) lie on one circle, so \(\angle BCP=\angle BAP\): these are inscribed angles subtending chord \(BP\). Hence \(\angle XPC=\angle ZPA\), and therefore \(\angle XYC=\angle ZYA\).

Since \(A,Y,C\) are collinear, equality \(\angle XYC=\angle ZYA\) means that rays \(YX\) and \(YZ\) lie on one line. Thus \(X,Y,Z\) are collinear.

Method comment. two circles with diameters \(PC\) and \(PA\) give equal angles at point \(Y\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since the tangent segments are equal, \(PB=PC\). Hence triangle \(PBC\) is isosceles, so the median \(PM\) to base \(BC\) is an altitude: \(PM\perp BC\).

Because \(PE\perp AC\), we have \(\angle PEC=90^\circ\). Also \(\angle PMC=90^\circ\). Therefore \(M,C,E,P\) lie on one circle.

In this circle, \(\angle MEP=\angle MCP\), since both angles subtend chord \(MP\). But \(CP\) is tangent to \(\Omega\) at \(C\), so by the tangent-chord theorem \(\angle MCP=\angle BAC\).

Since \(PE\perp AC\), we get \(\angle MEA=90^\circ-\angle MEP=90^\circ-\angle BAC\). Also \(\angle DAE=\angle BAC\), because \(D\in AB\) and \(E\in AC\). Thus \(\angle MEA+\angle DAE=90^\circ\), so \(ME\perp AD\).

Similarly, using the cyclic quadrilateral \(M,D,B,P\) and the tangent \(PB\), we obtain \(MD\perp AE\). Thus in triangle \(ADE\), the lines \(ME\) and \(MD\) are altitudes. Their intersection is \(M\), so \(M\) is the orthocenter of triangle \(ADE\).

Method comment. quadrilaterals \(MCEP\) and \(MDBP\) are cyclic, and the tangent converts angles at \(C\) and \(B\) into angles of the original triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(B,D,P,M\) lie on one circle, \(\angle BMP=\angle BDP\): both angles subtend chord \(BP\).

Points \(B,D,A\) are collinear, and points \(D,P,C\) are collinear. Hence \(\angle BDP\) is the oriented angle between lines \(AB\) and \(CD\).

Similarly, from the cyclic quadrilateral \(C,E,P,M\), we get \(\angle PME=\angle PCE\). But \(P,C,D\) are collinear and \(C,E,A\) are collinear, so \(\angle PCE\) is the oriented angle between \(CD\) and \(AC\).

Adding these two angles, we see that \(\angle BME\) equals the angle between \(AB\) and \(AC\), that is \(\angle BAC=66^\circ\).

But \(\angle BAE=\angle BAC\), since \(E\in AC\). Therefore \(\angle BME=\angle BAE\). By the converse criterion for equal inscribed angles, \(A,B,E,M\) lie on one circle.

Method comment. the second intersection of the two circles is a Miquel point and must also lie on the circle through \(A,B,E\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(PX\perp BC\) and \(PY\perp CA\), angles \(\angle PXC\) and \(\angle PYC\) are right angles. Therefore \(P,X,C,Y\) lie on one circle.

Similarly, \(PZ\perp AB\) and \(PY\perp CA\), so \(\angle PZA=\angle PYA=90^\circ\), and points \(P,Z,A,Y\) also lie on one circle.

From the first circle, \(\angle XYC=\angle XPC\). From the second circle, \(\angle ZYA=\angle ZPA\).

Now compare the right-hand sides. Since \(PX\perp BC\), \(\angle XPC=90^\circ-\angle BCP\). Since \(PZ\perp AB\), \(\angle ZPA=90^\circ-\angle BAP\).

Points \(A,B,C,P\) lie on one circle, so \(\angle BCP=\angle BAP\): these are inscribed angles subtending chord \(BP\). Hence \(\angle XPC=\angle ZPA\), and therefore \(\angle XYC=\angle ZYA\).

Since \(A,Y,C\) are collinear, equality \(\angle XYC=\angle ZYA\) means that rays \(YX\) and \(YZ\) lie on one line. Thus \(X,Y,Z\) are collinear.

Method comment. two circles with diameters \(PC\) and \(PA\) give equal angles at point \(Y\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since the tangent segments are equal, \(PB=PC\). Hence triangle \(PBC\) is isosceles, so the median \(PM\) to base \(BC\) is an altitude: \(PM\perp BC\).

Because \(PE\perp AC\), we have \(\angle PEC=90^\circ\). Also \(\angle PMC=90^\circ\). Therefore \(M,C,E,P\) lie on one circle.

In this circle, \(\angle MEP=\angle MCP\), since both angles subtend chord \(MP\). But \(CP\) is tangent to \(\Omega\) at \(C\), so by the tangent-chord theorem \(\angle MCP=\angle BAC\).

Since \(PE\perp AC\), we get \(\angle MEA=90^\circ-\angle MEP=90^\circ-\angle BAC\). Also \(\angle DAE=\angle BAC\), because \(D\in AB\) and \(E\in AC\). Thus \(\angle MEA+\angle DAE=90^\circ\), so \(ME\perp AD\).

Similarly, using the cyclic quadrilateral \(M,D,B,P\) and the tangent \(PB\), we obtain \(MD\perp AE\). Thus in triangle \(ADE\), the lines \(ME\) and \(MD\) are altitudes. Their intersection is \(M\), so \(M\) is the orthocenter of triangle \(ADE\).

Method comment. quadrilaterals \(MCEP\) and \(MDBP\) are cyclic, and the tangent converts angles at \(C\) and \(B\) into angles of the original triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since the tangent segments are equal, \(PB=PC\). Hence triangle \(PBC\) is isosceles, so the median \(PM\) to base \(BC\) is an altitude: \(PM\perp BC\).

Because \(PE\perp AC\), we have \(\angle PEC=90^\circ\). Also \(\angle PMC=90^\circ\). Therefore \(M,C,E,P\) lie on one circle.

In this circle, \(\angle MEP=\angle MCP\), since both angles subtend chord \(MP\). But \(CP\) is tangent to \(\Omega\) at \(C\), so by the tangent-chord theorem \(\angle MCP=\angle BAC\).

Since \(PE\perp AC\), we get \(\angle MEA=90^\circ-\angle MEP=90^\circ-\angle BAC\). Also \(\angle DAE=\angle BAC\), because \(D\in AB\) and \(E\in AC\). Thus \(\angle MEA+\angle DAE=90^\circ\), so \(ME\perp AD\).

Similarly, using the cyclic quadrilateral \(M,D,B,P\) and the tangent \(PB\), we obtain \(MD\perp AE\). Thus in triangle \(ADE\), the lines \(ME\) and \(MD\) are altitudes. Their intersection is \(M\), so \(M\) is the orthocenter of triangle \(ADE\).

Method comment. quadrilaterals \(MCEP\) and \(MDBP\) are cyclic, and the tangent converts angles at \(C\) and \(B\) into angles of the original triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(PX\perp BC\) and \(PY\perp CA\), angles \(\angle PXC\) and \(\angle PYC\) are right angles. Therefore \(P,X,C,Y\) lie on one circle.

Similarly, \(PZ\perp AB\) and \(PY\perp CA\), so \(\angle PZA=\angle PYA=90^\circ\), and points \(P,Z,A,Y\) also lie on one circle.

From the first circle, \(\angle XYC=\angle XPC\). From the second circle, \(\angle ZYA=\angle ZPA\).

Now compare the right-hand sides. Since \(PX\perp BC\), \(\angle XPC=90^\circ-\angle BCP\). Since \(PZ\perp AB\), \(\angle ZPA=90^\circ-\angle BAP\).

Points \(A,B,C,P\) lie on one circle, so \(\angle BCP=\angle BAP\): these are inscribed angles subtending chord \(BP\). Hence \(\angle XPC=\angle ZPA\), and therefore \(\angle XYC=\angle ZYA\).

Since \(A,Y,C\) are collinear, equality \(\angle XYC=\angle ZYA\) means that rays \(YX\) and \(YZ\) lie on one line. Thus \(X,Y,Z\) are collinear.

Method comment. two circles with diameters \(PC\) and \(PA\) give equal angles at point \(Y\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since the tangent segments are equal, \(PB=PC\). Hence triangle \(PBC\) is isosceles, so the median \(PM\) to base \(BC\) is an altitude: \(PM\perp BC\).

Because \(PE\perp AC\), we have \(\angle PEC=90^\circ\). Also \(\angle PMC=90^\circ\). Therefore \(M,C,E,P\) lie on one circle.

In this circle, \(\angle MEP=\angle MCP\), since both angles subtend chord \(MP\). But \(CP\) is tangent to \(\Omega\) at \(C\), so by the tangent-chord theorem \(\angle MCP=\angle BAC\).

Since \(PE\perp AC\), we get \(\angle MEA=90^\circ-\angle MEP=90^\circ-\angle BAC\). Also \(\angle DAE=\angle BAC\), because \(D\in AB\) and \(E\in AC\). Thus \(\angle MEA+\angle DAE=90^\circ\), so \(ME\perp AD\).

Similarly, using the cyclic quadrilateral \(M,D,B,P\) and the tangent \(PB\), we obtain \(MD\perp AE\). Thus in triangle \(ADE\), the lines \(ME\) and \(MD\) are altitudes. Their intersection is \(M\), so \(M\) is the orthocenter of triangle \(ADE\).

Method comment. quadrilaterals \(MCEP\) and \(MDBP\) are cyclic, and the tangent converts angles at \(C\) and \(B\) into angles of the original triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(B,D,P,M\) lie on one circle, \(\angle BMP=\angle BDP\): both angles subtend chord \(BP\).

Points \(B,D,A\) are collinear, and points \(D,P,C\) are collinear. Hence \(\angle BDP\) is the oriented angle between lines \(AB\) and \(CD\).

Similarly, from the cyclic quadrilateral \(C,E,P,M\), we get \(\angle PME=\angle PCE\). But \(P,C,D\) are collinear and \(C,E,A\) are collinear, so \(\angle PCE\) is the oriented angle between \(CD\) and \(AC\).

Adding these two angles, we see that \(\angle BME\) equals the angle between \(AB\) and \(AC\), that is \(\angle BAC=65^\circ\).

But \(\angle BAE=\angle BAC\), since \(E\in AC\). Therefore \(\angle BME=\angle BAE\). By the converse criterion for equal inscribed angles, \(A,B,E,M\) lie on one circle.

Method comment. the second intersection of the two circles is a Miquel point and must also lie on the circle through \(A,B,E\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(PX\perp BC\) and \(PY\perp CA\), angles \(\angle PXC\) and \(\angle PYC\) are right angles. Therefore \(P,X,C,Y\) lie on one circle.

Similarly, \(PZ\perp AB\) and \(PY\perp CA\), so \(\angle PZA=\angle PYA=90^\circ\), and points \(P,Z,A,Y\) also lie on one circle.

From the first circle, \(\angle XYC=\angle XPC\). From the second circle, \(\angle ZYA=\angle ZPA\).

Now compare the right-hand sides. Since \(PX\perp BC\), \(\angle XPC=90^\circ-\angle BCP\). Since \(PZ\perp AB\), \(\angle ZPA=90^\circ-\angle BAP\).

Points \(A,B,C,P\) lie on one circle, so \(\angle BCP=\angle BAP\): these are inscribed angles subtending chord \(BP\). Hence \(\angle XPC=\angle ZPA\), and therefore \(\angle XYC=\angle ZYA\).

Since \(A,Y,C\) are collinear, equality \(\angle XYC=\angle ZYA\) means that rays \(YX\) and \(YZ\) lie on one line. Thus \(X,Y,Z\) are collinear.

Method comment. two circles with diameters \(PC\) and \(PA\) give equal angles at point \(Y\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(B,D,P,M\) lie on one circle, \(\angle BMP=\angle BDP\): both angles subtend chord \(BP\).

Points \(B,D,A\) are collinear, and points \(D,P,C\) are collinear. Hence \(\angle BDP\) is the oriented angle between lines \(AB\) and \(CD\).

Similarly, from the cyclic quadrilateral \(C,E,P,M\), we get \(\angle PME=\angle PCE\). But \(P,C,D\) are collinear and \(C,E,A\) are collinear, so \(\angle PCE\) is the oriented angle between \(CD\) and \(AC\).

Adding these two angles, we see that \(\angle BME\) equals the angle between \(AB\) and \(AC\), that is \(\angle BAC=79^\circ\).

But \(\angle BAE=\angle BAC\), since \(E\in AC\). Therefore \(\angle BME=\angle BAE\). By the converse criterion for equal inscribed angles, \(A,B,E,M\) lie on one circle.

Method comment. the second intersection of the two circles is a Miquel point and must also lie on the circle through \(A,B,E\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(B,D,P,M\) lie on one circle, \(\angle BMP=\angle BDP\): both angles subtend chord \(BP\).

Points \(B,D,A\) are collinear, and points \(D,P,C\) are collinear. Hence \(\angle BDP\) is the oriented angle between lines \(AB\) and \(CD\).

Similarly, from the cyclic quadrilateral \(C,E,P,M\), we get \(\angle PME=\angle PCE\). But \(P,C,D\) are collinear and \(C,E,A\) are collinear, so \(\angle PCE\) is the oriented angle between \(CD\) and \(AC\).

Adding these two angles, we see that \(\angle BME\) equals the angle between \(AB\) and \(AC\), that is \(\angle BAC=43^\circ\).

But \(\angle BAE=\angle BAC\), since \(E\in AC\). Therefore \(\angle BME=\angle BAE\). By the converse criterion for equal inscribed angles, \(A,B,E,M\) lie on one circle.

Method comment. the second intersection of the two circles is a Miquel point and must also lie on the circle through \(A,B,E\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(PX\perp BC\) and \(PY\perp CA\), angles \(\angle PXC\) and \(\angle PYC\) are right angles. Therefore \(P,X,C,Y\) lie on one circle.

Similarly, \(PZ\perp AB\) and \(PY\perp CA\), so \(\angle PZA=\angle PYA=90^\circ\), and points \(P,Z,A,Y\) also lie on one circle.

From the first circle, \(\angle XYC=\angle XPC\). From the second circle, \(\angle ZYA=\angle ZPA\).

Now compare the right-hand sides. Since \(PX\perp BC\), \(\angle XPC=90^\circ-\angle BCP\). Since \(PZ\perp AB\), \(\angle ZPA=90^\circ-\angle BAP\).

Points \(A,B,C,P\) lie on one circle, so \(\angle BCP=\angle BAP\): these are inscribed angles subtending chord \(BP\). Hence \(\angle XPC=\angle ZPA\), and therefore \(\angle XYC=\angle ZYA\).

Since \(A,Y,C\) are collinear, equality \(\angle XYC=\angle ZYA\) means that rays \(YX\) and \(YZ\) lie on one line. Thus \(X,Y,Z\) are collinear.

Method comment. two circles with diameters \(PC\) and \(PA\) give equal angles at point \(Y\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since the tangent segments are equal, \(PB=PC\). Hence triangle \(PBC\) is isosceles, so the median \(PM\) to base \(BC\) is an altitude: \(PM\perp BC\).

Because \(PE\perp AC\), we have \(\angle PEC=90^\circ\). Also \(\angle PMC=90^\circ\). Therefore \(M,C,E,P\) lie on one circle.

In this circle, \(\angle MEP=\angle MCP\), since both angles subtend chord \(MP\). But \(CP\) is tangent to \(\Omega\) at \(C\), so by the tangent-chord theorem \(\angle MCP=\angle BAC\).

Since \(PE\perp AC\), we get \(\angle MEA=90^\circ-\angle MEP=90^\circ-\angle BAC\). Also \(\angle DAE=\angle BAC\), because \(D\in AB\) and \(E\in AC\). Thus \(\angle MEA+\angle DAE=90^\circ\), so \(ME\perp AD\).

Similarly, using the cyclic quadrilateral \(M,D,B,P\) and the tangent \(PB\), we obtain \(MD\perp AE\). Thus in triangle \(ADE\), the lines \(ME\) and \(MD\) are altitudes. Their intersection is \(M\), so \(M\) is the orthocenter of triangle \(ADE\).

Method comment. quadrilaterals \(MCEP\) and \(MDBP\) are cyclic, and the tangent converts angles at \(C\) and \(B\) into angles of the original triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(B,D,P,M\) lie on one circle, \(\angle BMP=\angle BDP\): both angles subtend chord \(BP\).

Points \(B,D,A\) are collinear, and points \(D,P,C\) are collinear. Hence \(\angle BDP\) is the oriented angle between lines \(AB\) and \(CD\).

Similarly, from the cyclic quadrilateral \(C,E,P,M\), we get \(\angle PME=\angle PCE\). But \(P,C,D\) are collinear and \(C,E,A\) are collinear, so \(\angle PCE\) is the oriented angle between \(CD\) and \(AC\).

Adding these two angles, we see that \(\angle BME\) equals the angle between \(AB\) and \(AC\), that is \(\angle BAC=64^\circ\).

But \(\angle BAE=\angle BAC\), since \(E\in AC\). Therefore \(\angle BME=\angle BAE\). By the converse criterion for equal inscribed angles, \(A,B,E,M\) lie on one circle.

Method comment. the second intersection of the two circles is a Miquel point and must also lie on the circle through \(A,B,E\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(PX\perp BC\) and \(PY\perp CA\), angles \(\angle PXC\) and \(\angle PYC\) are right angles. Therefore \(P,X,C,Y\) lie on one circle.

Similarly, \(PZ\perp AB\) and \(PY\perp CA\), so \(\angle PZA=\angle PYA=90^\circ\), and points \(P,Z,A,Y\) also lie on one circle.

From the first circle, \(\angle XYC=\angle XPC\). From the second circle, \(\angle ZYA=\angle ZPA\).

Now compare the right-hand sides. Since \(PX\perp BC\), \(\angle XPC=90^\circ-\angle BCP\). Since \(PZ\perp AB\), \(\angle ZPA=90^\circ-\angle BAP\).

Points \(A,B,C,P\) lie on one circle, so \(\angle BCP=\angle BAP\): these are inscribed angles subtending chord \(BP\). Hence \(\angle XPC=\angle ZPA\), and therefore \(\angle XYC=\angle ZYA\).

Since \(A,Y,C\) are collinear, equality \(\angle XYC=\angle ZYA\) means that rays \(YX\) and \(YZ\) lie on one line. Thus \(X,Y,Z\) are collinear.

Method comment. two circles with diameters \(PC\) and \(PA\) give equal angles at point \(Y\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since the tangent segments are equal, \(PB=PC\). Hence triangle \(PBC\) is isosceles, so the median \(PM\) to base \(BC\) is an altitude: \(PM\perp BC\).

Because \(PE\perp AC\), we have \(\angle PEC=90^\circ\). Also \(\angle PMC=90^\circ\). Therefore \(M,C,E,P\) lie on one circle.

In this circle, \(\angle MEP=\angle MCP\), since both angles subtend chord \(MP\). But \(CP\) is tangent to \(\Omega\) at \(C\), so by the tangent-chord theorem \(\angle MCP=\angle BAC\).

Since \(PE\perp AC\), we get \(\angle MEA=90^\circ-\angle MEP=90^\circ-\angle BAC\). Also \(\angle DAE=\angle BAC\), because \(D\in AB\) and \(E\in AC\). Thus \(\angle MEA+\angle DAE=90^\circ\), so \(ME\perp AD\).

Similarly, using the cyclic quadrilateral \(M,D,B,P\) and the tangent \(PB\), we obtain \(MD\perp AE\). Thus in triangle \(ADE\), the lines \(ME\) and \(MD\) are altitudes. Their intersection is \(M\), so \(M\) is the orthocenter of triangle \(ADE\).

Method comment. quadrilaterals \(MCEP\) and \(MDBP\) are cyclic, and the tangent converts angles at \(C\) and \(B\) into angles of the original triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(B,D,P,M\) lie on one circle, \(\angle BMP=\angle BDP\): both angles subtend chord \(BP\).

Points \(B,D,A\) are collinear, and points \(D,P,C\) are collinear. Hence \(\angle BDP\) is the oriented angle between lines \(AB\) and \(CD\).

Similarly, from the cyclic quadrilateral \(C,E,P,M\), we get \(\angle PME=\angle PCE\). But \(P,C,D\) are collinear and \(C,E,A\) are collinear, so \(\angle PCE\) is the oriented angle between \(CD\) and \(AC\).

Adding these two angles, we see that \(\angle BME\) equals the angle between \(AB\) and \(AC\), that is \(\angle BAC=42^\circ\).

But \(\angle BAE=\angle BAC\), since \(E\in AC\). Therefore \(\angle BME=\angle BAE\). By the converse criterion for equal inscribed angles, \(A,B,E,M\) lie on one circle.

Method comment. the second intersection of the two circles is a Miquel point and must also lie on the circle through \(A,B,E\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(PX\perp BC\) and \(PY\perp CA\), angles \(\angle PXC\) and \(\angle PYC\) are right angles. Therefore \(P,X,C,Y\) lie on one circle.

Similarly, \(PZ\perp AB\) and \(PY\perp CA\), so \(\angle PZA=\angle PYA=90^\circ\), and points \(P,Z,A,Y\) also lie on one circle.

From the first circle, \(\angle XYC=\angle XPC\). From the second circle, \(\angle ZYA=\angle ZPA\).

Now compare the right-hand sides. Since \(PX\perp BC\), \(\angle XPC=90^\circ-\angle BCP\). Since \(PZ\perp AB\), \(\angle ZPA=90^\circ-\angle BAP\).

Points \(A,B,C,P\) lie on one circle, so \(\angle BCP=\angle BAP\): these are inscribed angles subtending chord \(BP\). Hence \(\angle XPC=\angle ZPA\), and therefore \(\angle XYC=\angle ZYA\).

Since \(A,Y,C\) are collinear, equality \(\angle XYC=\angle ZYA\) means that rays \(YX\) and \(YZ\) lie on one line. Thus \(X,Y,Z\) are collinear.

Method comment. two circles with diameters \(PC\) and \(PA\) give equal angles at point \(Y\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since the tangent segments are equal, \(PB=PC\). Hence triangle \(PBC\) is isosceles, so the median \(PM\) to base \(BC\) is an altitude: \(PM\perp BC\).

Because \(PE\perp AC\), we have \(\angle PEC=90^\circ\). Also \(\angle PMC=90^\circ\). Therefore \(M,C,E,P\) lie on one circle.

In this circle, \(\angle MEP=\angle MCP\), since both angles subtend chord \(MP\). But \(CP\) is tangent to \(\Omega\) at \(C\), so by the tangent-chord theorem \(\angle MCP=\angle BAC\).

Since \(PE\perp AC\), we get \(\angle MEA=90^\circ-\angle MEP=90^\circ-\angle BAC\). Also \(\angle DAE=\angle BAC\), because \(D\in AB\) and \(E\in AC\). Thus \(\angle MEA+\angle DAE=90^\circ\), so \(ME\perp AD\).

Similarly, using the cyclic quadrilateral \(M,D,B,P\) and the tangent \(PB\), we obtain \(MD\perp AE\). Thus in triangle \(ADE\), the lines \(ME\) and \(MD\) are altitudes. Their intersection is \(M\), so \(M\) is the orthocenter of triangle \(ADE\).

Method comment. quadrilaterals \(MCEP\) and \(MDBP\) are cyclic, and the tangent converts angles at \(C\) and \(B\) into angles of the original triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(B,D,P,M\) lie on one circle, \(\angle BMP=\angle BDP\): both angles subtend chord \(BP\).

Points \(B,D,A\) are collinear, and points \(D,P,C\) are collinear. Hence \(\angle BDP\) is the oriented angle between lines \(AB\) and \(CD\).

Similarly, from the cyclic quadrilateral \(C,E,P,M\), we get \(\angle PME=\angle PCE\). But \(P,C,D\) are collinear and \(C,E,A\) are collinear, so \(\angle PCE\) is the oriented angle between \(CD\) and \(AC\).

Adding these two angles, we see that \(\angle BME\) equals the angle between \(AB\) and \(AC\), that is \(\angle BAC=63^\circ\).

But \(\angle BAE=\angle BAC\), since \(E\in AC\). Therefore \(\angle BME=\angle BAE\). By the converse criterion for equal inscribed angles, \(A,B,E,M\) lie on one circle.

Method comment. the second intersection of the two circles is a Miquel point and must also lie on the circle through \(A,B,E\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. By the tangent-chord theorem, \(\angle TAB=\angle ACB\), and \(\angle TAC=\angle ABC\). These two replacements are the main step of the problem.

Apply the sine rule in triangles \(TAB\) and \(TAC\). Since the angles at \(T\) are supplementary as angles on line \(BC\), after dividing the corresponding equalities we obtain \(\frac{TB}{TC}=\frac{\sin^2\angle ACB}{\sin^2\angle ABC}\).

In triangle \(ABC\), by the sine rule, \(\frac{AB}{AC}=\frac{\sin\angle ACB}{\sin\angle ABC}\). Hence \(\frac{TB}{TC}=\frac{AB^2}{AC^2}=\frac{64}{196}\). Therefore \(TB:TC=64:196\) in directed lengths.

Method comment. the tangent replaces angles at \(T\) by angles of the original triangle, after which the sine rule works Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. By the tangent-chord theorem, \(\angle TAB=\angle ACB\), and \(\angle TAC=\angle ABC\). These two replacements are the main step of the problem.

Apply the sine rule in triangles \(TAB\) and \(TAC\). Since the angles at \(T\) are supplementary as angles on line \(BC\), after dividing the corresponding equalities we obtain \(\frac{TB}{TC}=\frac{\sin^2\angle ACB}{\sin^2\angle ABC}\).

In triangle \(ABC\), by the sine rule, \(\frac{AB}{AC}=\frac{\sin\angle ACB}{\sin\angle ABC}\). Hence \(\frac{TB}{TC}=\frac{AB^2}{AC^2}=\frac{81}{256}\). Therefore \(TB:TC=81:256\) in directed lengths.

Method comment. the tangent replaces angles at \(T\) by angles of the original triangle, after which the sine rule works Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Let \(l\) be the tangent to \(\Omega\) at \(A\). By the statement, \(BD\parallel l\).

By the tangent-chord theorem, the angle between \(l\) and \(AB\) equals \(\angle ACB\). Since \(BD\parallel l\), we get \(\angle ABD=\angle ACB\).

Also \(D\in AC\), so \(\angle BAD=\angle BAC\). Therefore \(\triangle ABD\sim\triangle ACB\) by two angles.

From the similarity, \(\frac{AB}{AC}=\frac{AD}{AB}\). Multiplying gives \(AB^2=AC\cdot AD\), as required.

Method comment. the line parallel to the tangent transfers the tangent-chord theorem inside the triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(D\) is the foot of the altitude from \(B\), we have \(BD\perp AC\). Point \(H\) lies on this altitude, and \(A,D,C\) are collinear; hence \(DH\perp AD\), so \(\angle ADH=90^\circ\).

Similarly, \(E\) is the foot of the altitude from \(C\), so \(EH\perp AE\) and \(\angle AEH=90^\circ\).

Thus both \(D\) and \(E\) see segment \(AH\) under a right angle. Therefore \(D\) and \(E\) lie on the circle with diameter \(AH\). Hence \(A,D,H,E\) are concyclic.

In this circle, angles \(\angle DHE\) and \(\angle DAE\) subtend the same chord \(DE\). Therefore \(\angle DHE=\angle DAE=\angle BAC=38^\circ\).

Method comment. points \(D\) and \(E\) see segment \(AH\) under a right angle, so they should be placed on one circle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Let \(l\) be the tangent to \(\Omega\) at \(A\). By the statement, \(BD\parallel l\).

By the tangent-chord theorem, the angle between \(l\) and \(AB\) equals \(\angle ACB\). Since \(BD\parallel l\), we get \(\angle ABD=\angle ACB\).

Also \(D\in AC\), so \(\angle BAD=\angle BAC\). Therefore \(\triangle ABD\sim\triangle ACB\) by two angles.

From the similarity, \(\frac{AB}{AC}=\frac{AD}{AB}\). Multiplying gives \(AB^2=AC\cdot AD\), as required.

Method comment. the line parallel to the tangent transfers the tangent-chord theorem inside the triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(P,A,C\) are collinear, \(\angle PAB=\angle CAB\). Since \(P,D,B\) are collinear, \(\angle PDC=\angle CDB\).

Angles \(\angle CAB\) and \(\angle CDB\) are inscribed angles subtending the same chord \(CB\), so they are equal. Hence \(\angle PAB=\angle PDC\).

Similarly, \(\angle PBA=\angle DBA\), \(\angle PCD=\angle DCA\), and the inscribed angles \(\angle DBA\) and \(\angle DCA\) subtend chord \(DA\). Therefore \(\angle PBA=\angle PCD\).

By two angles, \(\triangle PAB\sim\triangle PDC\). From corresponding sides, \(\frac{PA}{PD}=\frac{AB}{CD}\). Hence \(PA=20\cdot\frac{9}{10}=18\).

Method comment. equal inscribed angles turn triangles \(PAB\) and \(PDC\) into similar triangles Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(D\) is the foot of the altitude from \(B\), we have \(BD\perp AC\). Point \(H\) lies on this altitude, and \(A,D,C\) are collinear; hence \(DH\perp AD\), so \(\angle ADH=90^\circ\).

Similarly, \(E\) is the foot of the altitude from \(C\), so \(EH\perp AE\) and \(\angle AEH=90^\circ\).

Thus both \(D\) and \(E\) see segment \(AH\) under a right angle. Therefore \(D\) and \(E\) lie on the circle with diameter \(AH\). Hence \(A,D,H,E\) are concyclic.

In this circle, angles \(\angle DHE\) and \(\angle DAE\) subtend the same chord \(DE\). Therefore \(\angle DHE=\angle DAE=\angle BAC=71^\circ\).

Method comment. points \(D\) and \(E\) see segment \(AH\) under a right angle, so they should be placed on one circle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. By the tangent-chord theorem, \(\angle TAB=\angle ACB\), and \(\angle TAC=\angle ABC\). These two replacements are the main step of the problem.

Apply the sine rule in triangles \(TAB\) and \(TAC\). Since the angles at \(T\) are supplementary as angles on line \(BC\), after dividing the corresponding equalities we obtain \(\frac{TB}{TC}=\frac{\sin^2\angle ACB}{\sin^2\angle ABC}\).

In triangle \(ABC\), by the sine rule, \(\frac{AB}{AC}=\frac{\sin\angle ACB}{\sin\angle ABC}\). Hence \(\frac{TB}{TC}=\frac{AB^2}{AC^2}=\frac{36}{289}\). Therefore \(TB:TC=36:289\) in directed lengths.

Method comment. the tangent replaces angles at \(T\) by angles of the original triangle, after which the sine rule works Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. By the tangent-chord theorem, \(\angle TAB=\angle ACB\), and \(\angle TAC=\angle ABC\). These two replacements are the main step of the problem.

Apply the sine rule in triangles \(TAB\) and \(TAC\). Since the angles at \(T\) are supplementary as angles on line \(BC\), after dividing the corresponding equalities we obtain \(\frac{TB}{TC}=\frac{\sin^2\angle ACB}{\sin^2\angle ABC}\).

In triangle \(ABC\), by the sine rule, \(\frac{AB}{AC}=\frac{\sin\angle ACB}{\sin\angle ABC}\). Hence \(\frac{TB}{TC}=\frac{AB^2}{AC^2}=\frac{49}{64}\). Therefore \(TB:TC=49:64\) in directed lengths.

Method comment. the tangent replaces angles at \(T\) by angles of the original triangle, after which the sine rule works Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(P,A,C\) are collinear, \(\angle PAB=\angle CAB\). Since \(P,D,B\) are collinear, \(\angle PDC=\angle CDB\).

Angles \(\angle CAB\) and \(\angle CDB\) are inscribed angles subtending the same chord \(CB\), so they are equal. Hence \(\angle PAB=\angle PDC\).

Similarly, \(\angle PBA=\angle DBA\), \(\angle PCD=\angle DCA\), and the inscribed angles \(\angle DBA\) and \(\angle DCA\) subtend chord \(DA\). Therefore \(\angle PBA=\angle PCD\).

By two angles, \(\triangle PAB\sim\triangle PDC\). From corresponding sides, \(\frac{PA}{PD}=\frac{AB}{CD}\). Hence \(PA=26\cdot\frac{5}{13}=10\).

Method comment. equal inscribed angles turn triangles \(PAB\) and \(PDC\) into similar triangles Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(P,A,C\) are collinear, \(\angle PAB=\angle CAB\). Since \(P,D,B\) are collinear, \(\angle PDC=\angle CDB\).

Angles \(\angle CAB\) and \(\angle CDB\) are inscribed angles subtending the same chord \(CB\), so they are equal. Hence \(\angle PAB=\angle PDC\).

Similarly, \(\angle PBA=\angle DBA\), \(\angle PCD=\angle DCA\), and the inscribed angles \(\angle DBA\) and \(\angle DCA\) subtend chord \(DA\). Therefore \(\angle PBA=\angle PCD\).

By two angles, \(\triangle PAB\sim\triangle PDC\). From corresponding sides, \(\frac{PA}{PD}=\frac{AB}{CD}\). Hence \(PA=21\cdot\frac{6}{7}=18\).

Method comment. equal inscribed angles turn triangles \(PAB\) and \(PDC\) into similar triangles Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. For the orthocenter of an acute triangle, \(\angle AHB=180^\circ-\angle ACB\).

Under reflection across line \(AB\), points \(A\) and \(B\) remain fixed, and \(H\) maps to \(H_b\). Hence \(\angle AH_bB=\angle AHB=180^\circ-\angle ACB\).

Thus the opposite angles of quadrilateral \(A,B,C,H_b\) sum to \(180^\circ\). Therefore \(A,B,C,H_b\) are concyclic, so \(H_b\) lies on the circumcircle of triangle \(ABC\).

In the same way, reflecting \(H\) across \(AC\), we obtain \(H_c\) on the same circumcircle.

Since \(B,C,H_b,H_c\) lie on one circle, namely the circumcircle of triangle \(ABC\), quadrilateral \(B,C,H_c,H_b\) is cyclic.

Method comment. reflection preserves the angle with the mirror line, and the angle at the orthocenter supplements an angle of the triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(P,A,C\) are collinear, \(\angle PAB=\angle CAB\). Since \(P,D,B\) are collinear, \(\angle PDC=\angle CDB\).

Angles \(\angle CAB\) and \(\angle CDB\) are inscribed angles subtending the same chord \(CB\), so they are equal. Hence \(\angle PAB=\angle PDC\).

Similarly, \(\angle PBA=\angle DBA\), \(\angle PCD=\angle DCA\), and the inscribed angles \(\angle DBA\) and \(\angle DCA\) subtend chord \(DA\). Therefore \(\angle PBA=\angle PCD\).

By two angles, \(\triangle PAB\sim\triangle PDC\). From corresponding sides, \(\frac{PA}{PD}=\frac{AB}{CD}\). Hence \(PA=65\cdot\frac{8}{13}=40\).

Method comment. equal inscribed angles turn triangles \(PAB\) and \(PDC\) into similar triangles Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Let \(l\) be the tangent to \(\Omega\) at \(A\). By the statement, \(BD\parallel l\).

By the tangent-chord theorem, the angle between \(l\) and \(AB\) equals \(\angle ACB\). Since \(BD\parallel l\), we get \(\angle ABD=\angle ACB\).

Also \(D\in AC\), so \(\angle BAD=\angle BAC\). Therefore \(\triangle ABD\sim\triangle ACB\) by two angles.

From the similarity, \(\frac{AB}{AC}=\frac{AD}{AB}\). Multiplying gives \(AB^2=AC\cdot AD\), as required.

Method comment. the line parallel to the tangent transfers the tangent-chord theorem inside the triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Draw tangents \(t_1\) and \(t_2\) at \(A\) to circles \(\omega_1\) and \(\omega_2\). The required angle is the acute angle between \(t_1\) and \(t_2\).

By the tangent-chord theorem, the angle between \(t_1\) and \(AB\) equals the inscribed angle \(\angle ACB=83^\circ\). Similarly, the angle between \(t_2\) and \(AB\) equals \(\angle ADB=50^\circ\).

Since \(C\) and \(D\) lie on the same side of \(AB\), the corresponding tangents are compared on the same side of line \(AB\). Hence the acute angle between the circles is the difference: \(83^\circ-50^\circ=33^\circ\).

Method comment. the angle between circles equals the angle between tangents, and each tangent can be replaced by an inscribed angle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. By the tangent-chord theorem, \(\angle TAB=\angle ACB\), and \(\angle TAC=\angle ABC\). These two replacements are the main step of the problem.

Apply the sine rule in triangles \(TAB\) and \(TAC\). Since the angles at \(T\) are supplementary as angles on line \(BC\), after dividing the corresponding equalities we obtain \(\frac{TB}{TC}=\frac{\sin^2\angle ACB}{\sin^2\angle ABC}\).

In triangle \(ABC\), by the sine rule, \(\frac{AB}{AC}=\frac{\sin\angle ACB}{\sin\angle ABC}\). Hence \(\frac{TB}{TC}=\frac{AB^2}{AC^2}=\frac{196}{121}\). Therefore \(TB:TC=196:121\) in directed lengths.

Method comment. the tangent replaces angles at \(T\) by angles of the original triangle, after which the sine rule works Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Draw tangents \(t_1\) and \(t_2\) at \(A\) to circles \(\omega_1\) and \(\omega_2\). The required angle is the acute angle between \(t_1\) and \(t_2\).

By the tangent-chord theorem, the angle between \(t_1\) and \(AB\) equals the inscribed angle \(\angle ACB=58^\circ\). Similarly, the angle between \(t_2\) and \(AB\) equals \(\angle ADB=40^\circ\).

Since \(C\) and \(D\) lie on the same side of \(AB\), the corresponding tangents are compared on the same side of line \(AB\). Hence the acute angle between the circles is the difference: \(58^\circ-40^\circ=18^\circ\).

Method comment. the angle between circles equals the angle between tangents, and each tangent can be replaced by an inscribed angle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(P,A,C\) are collinear, \(\angle PAB=\angle CAB\). Since \(P,D,B\) are collinear, \(\angle PDC=\angle CDB\).

Angles \(\angle CAB\) and \(\angle CDB\) are inscribed angles subtending the same chord \(CB\), so they are equal. Hence \(\angle PAB=\angle PDC\).

Similarly, \(\angle PBA=\angle DBA\), \(\angle PCD=\angle DCA\), and the inscribed angles \(\angle DBA\) and \(\angle DCA\) subtend chord \(DA\). Therefore \(\angle PBA=\angle PCD\).

By two angles, \(\triangle PAB\sim\triangle PDC\). From corresponding sides, \(\frac{PA}{PD}=\frac{AB}{CD}\). Hence \(PA=20\cdot\frac{5}{10}=10\).

Method comment. equal inscribed angles turn triangles \(PAB\) and \(PDC\) into similar triangles Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Let \(l\) be the tangent to \(\Omega\) at \(A\). By the statement, \(BD\parallel l\).

By the tangent-chord theorem, the angle between \(l\) and \(AB\) equals \(\angle ACB\). Since \(BD\parallel l\), we get \(\angle ABD=\angle ACB\).

Also \(D\in AC\), so \(\angle BAD=\angle BAC\). Therefore \(\triangle ABD\sim\triangle ACB\) by two angles.

From the similarity, \(\frac{AB}{AC}=\frac{AD}{AB}\). Multiplying gives \(AB^2=AC\cdot AD\), as required.

Method comment. the line parallel to the tangent transfers the tangent-chord theorem inside the triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(PX\perp BC\) and \(PY\perp CA\), angles \(\angle PXC\) and \(\angle PYC\) are right angles. Therefore \(P,X,C,Y\) lie on one circle.

Similarly, \(PZ\perp AB\) and \(PY\perp CA\), so \(\angle PZA=\angle PYA=90^\circ\), and points \(P,Z,A,Y\) also lie on one circle.

From the first circle, \(\angle XYC=\angle XPC\). From the second circle, \(\angle ZYA=\angle ZPA\).

Now compare the right-hand sides. Since \(PX\perp BC\), \(\angle XPC=90^\circ-\angle BCP\). Since \(PZ\perp AB\), \(\angle ZPA=90^\circ-\angle BAP\).

Points \(A,B,C,P\) lie on one circle, so \(\angle BCP=\angle BAP\): these are inscribed angles subtending chord \(BP\). Hence \(\angle XPC=\angle ZPA\), and therefore \(\angle XYC=\angle ZYA\).

Since \(A,Y,C\) are collinear, equality \(\angle XYC=\angle ZYA\) means that rays \(YX\) and \(YZ\) lie on one line. Thus \(X,Y,Z\) are collinear.

Method comment. two circles with diameters \(PC\) and \(PA\) give equal angles at point \(Y\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since the tangent segments are equal, \(PB=PC\). Hence triangle \(PBC\) is isosceles, so the median \(PM\) to base \(BC\) is an altitude: \(PM\perp BC\).

Because \(PE\perp AC\), we have \(\angle PEC=90^\circ\). Also \(\angle PMC=90^\circ\). Therefore \(M,C,E,P\) lie on one circle.

In this circle, \(\angle MEP=\angle MCP\), since both angles subtend chord \(MP\). But \(CP\) is tangent to \(\Omega\) at \(C\), so by the tangent-chord theorem \(\angle MCP=\angle BAC\).

Since \(PE\perp AC\), we get \(\angle MEA=90^\circ-\angle MEP=90^\circ-\angle BAC\). Also \(\angle DAE=\angle BAC\), because \(D\in AB\) and \(E\in AC\). Thus \(\angle MEA+\angle DAE=90^\circ\), so \(ME\perp AD\).

Similarly, using the cyclic quadrilateral \(M,D,B,P\) and the tangent \(PB\), we obtain \(MD\perp AE\). Thus in triangle \(ADE\), the lines \(ME\) and \(MD\) are altitudes. Their intersection is \(M\), so \(M\) is the orthocenter of triangle \(ADE\).

Method comment. quadrilaterals \(MCEP\) and \(MDBP\) are cyclic, and the tangent converts angles at \(C\) and \(B\) into angles of the original triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Draw tangents \(t_1\) and \(t_2\) at \(A\) to circles \(\omega_1\) and \(\omega_2\). The required angle is the acute angle between \(t_1\) and \(t_2\).

By the tangent-chord theorem, the angle between \(t_1\) and \(AB\) equals the inscribed angle \(\angle ACB=83^\circ\). Similarly, the angle between \(t_2\) and \(AB\) equals \(\angle ADB=51^\circ\).

Since \(C\) and \(D\) lie on the same side of \(AB\), the corresponding tangents are compared on the same side of line \(AB\). Hence the acute angle between the circles is the difference: \(83^\circ-51^\circ=32^\circ\).

Method comment. the angle between circles equals the angle between tangents, and each tangent can be replaced by an inscribed angle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(D\) is the foot of the altitude from \(B\), we have \(BD\perp AC\). Point \(H\) lies on this altitude, and \(A,D,C\) are collinear; hence \(DH\perp AD\), so \(\angle ADH=90^\circ\).

Similarly, \(E\) is the foot of the altitude from \(C\), so \(EH\perp AE\) and \(\angle AEH=90^\circ\).

Thus both \(D\) and \(E\) see segment \(AH\) under a right angle. Therefore \(D\) and \(E\) lie on the circle with diameter \(AH\). Hence \(A,D,H,E\) are concyclic.

In this circle, angles \(\angle DHE\) and \(\angle DAE\) subtend the same chord \(DE\). Therefore \(\angle DHE=\angle DAE=\angle BAC=64^\circ\).

Method comment. points \(D\) and \(E\) see segment \(AH\) under a right angle, so they should be placed on one circle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Draw tangents \(t_1\) and \(t_2\) at \(A\) to circles \(\omega_1\) and \(\omega_2\). The required angle is the acute angle between \(t_1\) and \(t_2\).

By the tangent-chord theorem, the angle between \(t_1\) and \(AB\) equals the inscribed angle \(\angle ACB=58^\circ\). Similarly, the angle between \(t_2\) and \(AB\) equals \(\angle ADB=41^\circ\).

Since \(C\) and \(D\) lie on the same side of \(AB\), the corresponding tangents are compared on the same side of line \(AB\). Hence the acute angle between the circles is the difference: \(58^\circ-41^\circ=17^\circ\).

Method comment. the angle between circles equals the angle between tangents, and each tangent can be replaced by an inscribed angle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. By the tangent-chord theorem, \(\angle TAB=\angle ACB\), and \(\angle TAC=\angle ABC\). These two replacements are the main step of the problem.

Apply the sine rule in triangles \(TAB\) and \(TAC\). Since the angles at \(T\) are supplementary as angles on line \(BC\), after dividing the corresponding equalities we obtain \(\frac{TB}{TC}=\frac{\sin^2\angle ACB}{\sin^2\angle ABC}\).

In triangle \(ABC\), by the sine rule, \(\frac{AB}{AC}=\frac{\sin\angle ACB}{\sin\angle ABC}\). Hence \(\frac{TB}{TC}=\frac{AB^2}{AC^2}=\frac{196}{324}\). Therefore \(TB:TC=196:324\) in directed lengths.

Method comment. the tangent replaces angles at \(T\) by angles of the original triangle, after which the sine rule works Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(B,D,P,M\) lie on one circle, \(\angle BMP=\angle BDP\): both angles subtend chord \(BP\).

Points \(B,D,A\) are collinear, and points \(D,P,C\) are collinear. Hence \(\angle BDP\) is the oriented angle between lines \(AB\) and \(CD\).

Similarly, from the cyclic quadrilateral \(C,E,P,M\), we get \(\angle PME=\angle PCE\). But \(P,C,D\) are collinear and \(C,E,A\) are collinear, so \(\angle PCE\) is the oriented angle between \(CD\) and \(AC\).

Adding these two angles, we see that \(\angle BME\) equals the angle between \(AB\) and \(AC\), that is \(\angle BAC=73^\circ\).

But \(\angle BAE=\angle BAC\), since \(E\in AC\). Therefore \(\angle BME=\angle BAE\). By the converse criterion for equal inscribed angles, \(A,B,E,M\) lie on one circle.

Method comment. the second intersection of the two circles is a Miquel point and must also lie on the circle through \(A,B,E\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(P,A,C\) are collinear, \(\angle PAB=\angle CAB\). Since \(P,D,B\) are collinear, \(\angle PDC=\angle CDB\).

Angles \(\angle CAB\) and \(\angle CDB\) are inscribed angles subtending the same chord \(CB\), so they are equal. Hence \(\angle PAB=\angle PDC\).

Similarly, \(\angle PBA=\angle DBA\), \(\angle PCD=\angle DCA\), and the inscribed angles \(\angle DBA\) and \(\angle DCA\) subtend chord \(DA\). Therefore \(\angle PBA=\angle PCD\).

By two angles, \(\triangle PAB\sim\triangle PDC\). From corresponding sides, \(\frac{PA}{PD}=\frac{AB}{CD}\). Hence \(PA=30\cdot\frac{6}{10}=18\).

Method comment. equal inscribed angles turn triangles \(PAB\) and \(PDC\) into similar triangles Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(D\) is the foot of the altitude from \(B\), we have \(BD\perp AC\). Point \(H\) lies on this altitude, and \(A,D,C\) are collinear; hence \(DH\perp AD\), so \(\angle ADH=90^\circ\).

Similarly, \(E\) is the foot of the altitude from \(C\), so \(EH\perp AE\) and \(\angle AEH=90^\circ\).

Thus both \(D\) and \(E\) see segment \(AH\) under a right angle. Therefore \(D\) and \(E\) lie on the circle with diameter \(AH\). Hence \(A,D,H,E\) are concyclic.

In this circle, angles \(\angle DHE\) and \(\angle DAE\) subtend the same chord \(DE\). Therefore \(\angle DHE=\angle DAE=\angle BAC=58^\circ\).

Method comment. points \(D\) and \(E\) see segment \(AH\) under a right angle, so they should be placed on one circle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since the tangent segments are equal, \(PB=PC\). Hence triangle \(PBC\) is isosceles, so the median \(PM\) to base \(BC\) is an altitude: \(PM\perp BC\).

Because \(PE\perp AC\), we have \(\angle PEC=90^\circ\). Also \(\angle PMC=90^\circ\). Therefore \(M,C,E,P\) lie on one circle.

In this circle, \(\angle MEP=\angle MCP\), since both angles subtend chord \(MP\). But \(CP\) is tangent to \(\Omega\) at \(C\), so by the tangent-chord theorem \(\angle MCP=\angle BAC\).

Since \(PE\perp AC\), we get \(\angle MEA=90^\circ-\angle MEP=90^\circ-\angle BAC\). Also \(\angle DAE=\angle BAC\), because \(D\in AB\) and \(E\in AC\). Thus \(\angle MEA+\angle DAE=90^\circ\), so \(ME\perp AD\).

Similarly, using the cyclic quadrilateral \(M,D,B,P\) and the tangent \(PB\), we obtain \(MD\perp AE\). Thus in triangle \(ADE\), the lines \(ME\) and \(MD\) are altitudes. Their intersection is \(M\), so \(M\) is the orthocenter of triangle \(ADE\).

Method comment. quadrilaterals \(MCEP\) and \(MDBP\) are cyclic, and the tangent converts angles at \(C\) and \(B\) into angles of the original triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. By the tangent-chord theorem, \(\angle TAB=\angle ACB\), and \(\angle TAC=\angle ABC\). These two replacements are the main step of the problem.

Apply the sine rule in triangles \(TAB\) and \(TAC\). Since the angles at \(T\) are supplementary as angles on line \(BC\), after dividing the corresponding equalities we obtain \(\frac{TB}{TC}=\frac{\sin^2\angle ACB}{\sin^2\angle ABC}\).

In triangle \(ABC\), by the sine rule, \(\frac{AB}{AC}=\frac{\sin\angle ACB}{\sin\angle ABC}\). Hence \(\frac{TB}{TC}=\frac{AB^2}{AC^2}=\frac{100}{289}\). Therefore \(TB:TC=100:289\) in directed lengths.

Method comment. the tangent replaces angles at \(T\) by angles of the original triangle, after which the sine rule works Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(D\) is the foot of the altitude from \(B\), we have \(BD\perp AC\). Point \(H\) lies on this altitude, and \(A,D,C\) are collinear; hence \(DH\perp AD\), so \(\angle ADH=90^\circ\).

Similarly, \(E\) is the foot of the altitude from \(C\), so \(EH\perp AE\) and \(\angle AEH=90^\circ\).

Thus both \(D\) and \(E\) see segment \(AH\) under a right angle. Therefore \(D\) and \(E\) lie on the circle with diameter \(AH\). Hence \(A,D,H,E\) are concyclic.

In this circle, angles \(\angle DHE\) and \(\angle DAE\) subtend the same chord \(DE\). Therefore \(\angle DHE=\angle DAE=\angle BAC=91^\circ\).

Method comment. points \(D\) and \(E\) see segment \(AH\) under a right angle, so they should be placed on one circle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(P,A,C\) are collinear, \(\angle PAB=\angle CAB\). Since \(P,D,B\) are collinear, \(\angle PDC=\angle CDB\).

Angles \(\angle CAB\) and \(\angle CDB\) are inscribed angles subtending the same chord \(CB\), so they are equal. Hence \(\angle PAB=\angle PDC\).

Similarly, \(\angle PBA=\angle DBA\), \(\angle PCD=\angle DCA\), and the inscribed angles \(\angle DBA\) and \(\angle DCA\) subtend chord \(DA\). Therefore \(\angle PBA=\angle PCD\).

By two angles, \(\triangle PAB\sim\triangle PDC\). From corresponding sides, \(\frac{PA}{PD}=\frac{AB}{CD}\). Hence \(PA=28\cdot\frac{11}{7}=44\).

Method comment. equal inscribed angles turn triangles \(PAB\) and \(PDC\) into similar triangles Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. By the tangent-chord theorem, \(\angle TAB=\angle ACB\), and \(\angle TAC=\angle ABC\). These two replacements are the main step of the problem.

Apply the sine rule in triangles \(TAB\) and \(TAC\). Since the angles at \(T\) are supplementary as angles on line \(BC\), after dividing the corresponding equalities we obtain \(\frac{TB}{TC}=\frac{\sin^2\angle ACB}{\sin^2\angle ABC}\).

In triangle \(ABC\), by the sine rule, \(\frac{AB}{AC}=\frac{\sin\angle ACB}{\sin\angle ABC}\). Hence \(\frac{TB}{TC}=\frac{AB^2}{AC^2}=\frac{169}{144}\). Therefore \(TB:TC=169:144\) in directed lengths.

Method comment. the tangent replaces angles at \(T\) by angles of the original triangle, after which the sine rule works Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(P,A,C\) are collinear, \(\angle PAB=\angle CAB\). Since \(P,D,B\) are collinear, \(\angle PDC=\angle CDB\).

Angles \(\angle CAB\) and \(\angle CDB\) are inscribed angles subtending the same chord \(CB\), so they are equal. Hence \(\angle PAB=\angle PDC\).

Similarly, \(\angle PBA=\angle DBA\), \(\angle PCD=\angle DCA\), and the inscribed angles \(\angle DBA\) and \(\angle DCA\) subtend chord \(DA\). Therefore \(\angle PBA=\angle PCD\).

By two angles, \(\triangle PAB\sim\triangle PDC\). From corresponding sides, \(\frac{PA}{PD}=\frac{AB}{CD}\). Hence \(PA=26\cdot\frac{5}{13}=10\).

Method comment. equal inscribed angles turn triangles \(PAB\) and \(PDC\) into similar triangles Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since the tangent segments are equal, \(PB=PC\). Hence triangle \(PBC\) is isosceles, so the median \(PM\) to base \(BC\) is an altitude: \(PM\perp BC\).

Because \(PE\perp AC\), we have \(\angle PEC=90^\circ\). Also \(\angle PMC=90^\circ\). Therefore \(M,C,E,P\) lie on one circle.

In this circle, \(\angle MEP=\angle MCP\), since both angles subtend chord \(MP\). But \(CP\) is tangent to \(\Omega\) at \(C\), so by the tangent-chord theorem \(\angle MCP=\angle BAC\).

Since \(PE\perp AC\), we get \(\angle MEA=90^\circ-\angle MEP=90^\circ-\angle BAC\). Also \(\angle DAE=\angle BAC\), because \(D\in AB\) and \(E\in AC\). Thus \(\angle MEA+\angle DAE=90^\circ\), so \(ME\perp AD\).

Similarly, using the cyclic quadrilateral \(M,D,B,P\) and the tangent \(PB\), we obtain \(MD\perp AE\). Thus in triangle \(ADE\), the lines \(ME\) and \(MD\) are altitudes. Their intersection is \(M\), so \(M\) is the orthocenter of triangle \(ADE\).

Method comment. quadrilaterals \(MCEP\) and \(MDBP\) are cyclic, and the tangent converts angles at \(C\) and \(B\) into angles of the original triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(D\) is the foot of the altitude from \(B\), we have \(BD\perp AC\). Point \(H\) lies on this altitude, and \(A,D,C\) are collinear; hence \(DH\perp AD\), so \(\angle ADH=90^\circ\).

Similarly, \(E\) is the foot of the altitude from \(C\), so \(EH\perp AE\) and \(\angle AEH=90^\circ\).

Thus both \(D\) and \(E\) see segment \(AH\) under a right angle. Therefore \(D\) and \(E\) lie on the circle with diameter \(AH\). Hence \(A,D,H,E\) are concyclic.

In this circle, angles \(\angle DHE\) and \(\angle DAE\) subtend the same chord \(DE\). Therefore \(\angle DHE=\angle DAE=\angle BAC=85^\circ\).

Method comment. points \(D\) and \(E\) see segment \(AH\) under a right angle, so they should be placed on one circle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. By the tangent-chord theorem, \(\angle TAB=\angle ACB\), and \(\angle TAC=\angle ABC\). These two replacements are the main step of the problem.

Apply the sine rule in triangles \(TAB\) and \(TAC\). Since the angles at \(T\) are supplementary as angles on line \(BC\), after dividing the corresponding equalities we obtain \(\frac{TB}{TC}=\frac{\sin^2\angle ACB}{\sin^2\angle ABC}\).

In triangle \(ABC\), by the sine rule, \(\frac{AB}{AC}=\frac{\sin\angle ACB}{\sin\angle ABC}\). Hence \(\frac{TB}{TC}=\frac{AB^2}{AC^2}=\frac{64}{81}\). Therefore \(TB:TC=64:81\) in directed lengths.

Method comment. the tangent replaces angles at \(T\) by angles of the original triangle, after which the sine rule works Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Let \(l\) be the tangent to \(\Omega\) at \(A\). By the statement, \(BD\parallel l\).

By the tangent-chord theorem, the angle between \(l\) and \(AB\) equals \(\angle ACB\). Since \(BD\parallel l\), we get \(\angle ABD=\angle ACB\).

Also \(D\in AC\), so \(\angle BAD=\angle BAC\). Therefore \(\triangle ABD\sim\triangle ACB\) by two angles.

From the similarity, \(\frac{AB}{AC}=\frac{AD}{AB}\). Multiplying gives \(AB^2=AC\cdot AD\), as required.

Method comment. the line parallel to the tangent transfers the tangent-chord theorem inside the triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(P,A,C\) are collinear, \(\angle PAB=\angle CAB\). Since \(P,D,B\) are collinear, \(\angle PDC=\angle CDB\).

Angles \(\angle CAB\) and \(\angle CDB\) are inscribed angles subtending the same chord \(CB\), so they are equal. Hence \(\angle PAB=\angle PDC\).

Similarly, \(\angle PBA=\angle DBA\), \(\angle PCD=\angle DCA\), and the inscribed angles \(\angle DBA\) and \(\angle DCA\) subtend chord \(DA\). Therefore \(\angle PBA=\angle PCD\).

By two angles, \(\triangle PAB\sim\triangle PDC\). From corresponding sides, \(\frac{PA}{PD}=\frac{AB}{CD}\). Hence \(PA=30\cdot\frac{10}{10}=30\).

Method comment. equal inscribed angles turn triangles \(PAB\) and \(PDC\) into similar triangles Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(P,A,C\) are collinear, \(\angle PAB=\angle CAB\). Since \(P,D,B\) are collinear, \(\angle PDC=\angle CDB\).

Angles \(\angle CAB\) and \(\angle CDB\) are inscribed angles subtending the same chord \(CB\), so they are equal. Hence \(\angle PAB=\angle PDC\).

Similarly, \(\angle PBA=\angle DBA\), \(\angle PCD=\angle DCA\), and the inscribed angles \(\angle DBA\) and \(\angle DCA\) subtend chord \(DA\). Therefore \(\angle PBA=\angle PCD\).

By two angles, \(\triangle PAB\sim\triangle PDC\). From corresponding sides, \(\frac{PA}{PD}=\frac{AB}{CD}\). Hence \(PA=52\cdot\frac{11}{13}=44\).

Method comment. equal inscribed angles turn triangles \(PAB\) and \(PDC\) into similar triangles Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. By the tangent-chord theorem, \(\angle TAB=\angle ACB\), and \(\angle TAC=\angle ABC\). These two replacements are the main step of the problem.

Apply the sine rule in triangles \(TAB\) and \(TAC\). Since the angles at \(T\) are supplementary as angles on line \(BC\), after dividing the corresponding equalities we obtain \(\frac{TB}{TC}=\frac{\sin^2\angle ACB}{\sin^2\angle ABC}\).

In triangle \(ABC\), by the sine rule, \(\frac{AB}{AC}=\frac{\sin\angle ACB}{\sin\angle ABC}\). Hence \(\frac{TB}{TC}=\frac{AB^2}{AC^2}=\frac{144}{289}\). Therefore \(TB:TC=144:289\) in directed lengths.

Method comment. the tangent replaces angles at \(T\) by angles of the original triangle, after which the sine rule works Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(D\) is the foot of the altitude from \(B\), we have \(BD\perp AC\). Point \(H\) lies on this altitude, and \(A,D,C\) are collinear; hence \(DH\perp AD\), so \(\angle ADH=90^\circ\).

Similarly, \(E\) is the foot of the altitude from \(C\), so \(EH\perp AE\) and \(\angle AEH=90^\circ\).

Thus both \(D\) and \(E\) see segment \(AH\) under a right angle. Therefore \(D\) and \(E\) lie on the circle with diameter \(AH\). Hence \(A,D,H,E\) are concyclic.

In this circle, angles \(\angle DHE\) and \(\angle DAE\) subtend the same chord \(DE\). Therefore \(\angle DHE=\angle DAE=\angle BAC=90^\circ\).

Method comment. points \(D\) and \(E\) see segment \(AH\) under a right angle, so they should be placed on one circle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Draw tangents \(t_1\) and \(t_2\) at \(A\) to circles \(\omega_1\) and \(\omega_2\). The required angle is the acute angle between \(t_1\) and \(t_2\).

By the tangent-chord theorem, the angle between \(t_1\) and \(AB\) equals the inscribed angle \(\angle ACB=83^\circ\). Similarly, the angle between \(t_2\) and \(AB\) equals \(\angle ADB=54^\circ\).

Since \(C\) and \(D\) lie on the same side of \(AB\), the corresponding tangents are compared on the same side of line \(AB\). Hence the acute angle between the circles is the difference: \(83^\circ-54^\circ=29^\circ\).

Method comment. the angle between circles equals the angle between tangents, and each tangent can be replaced by an inscribed angle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since the tangent segments are equal, \(PB=PC\). Hence triangle \(PBC\) is isosceles, so the median \(PM\) to base \(BC\) is an altitude: \(PM\perp BC\).

Because \(PE\perp AC\), we have \(\angle PEC=90^\circ\). Also \(\angle PMC=90^\circ\). Therefore \(M,C,E,P\) lie on one circle.

In this circle, \(\angle MEP=\angle MCP\), since both angles subtend chord \(MP\). But \(CP\) is tangent to \(\Omega\) at \(C\), so by the tangent-chord theorem \(\angle MCP=\angle BAC\).

Since \(PE\perp AC\), we get \(\angle MEA=90^\circ-\angle MEP=90^\circ-\angle BAC\). Also \(\angle DAE=\angle BAC\), because \(D\in AB\) and \(E\in AC\). Thus \(\angle MEA+\angle DAE=90^\circ\), so \(ME\perp AD\).

Similarly, using the cyclic quadrilateral \(M,D,B,P\) and the tangent \(PB\), we obtain \(MD\perp AE\). Thus in triangle \(ADE\), the lines \(ME\) and \(MD\) are altitudes. Their intersection is \(M\), so \(M\) is the orthocenter of triangle \(ADE\).

Method comment. quadrilaterals \(MCEP\) and \(MDBP\) are cyclic, and the tangent converts angles at \(C\) and \(B\) into angles of the original triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(P,A,C\) are collinear, \(\angle PAB=\angle CAB\). Since \(P,D,B\) are collinear, \(\angle PDC=\angle CDB\).

Angles \(\angle CAB\) and \(\angle CDB\) are inscribed angles subtending the same chord \(CB\), so they are equal. Hence \(\angle PAB=\angle PDC\).

Similarly, \(\angle PBA=\angle DBA\), \(\angle PCD=\angle DCA\), and the inscribed angles \(\angle DBA\) and \(\angle DCA\) subtend chord \(DA\). Therefore \(\angle PBA=\angle PCD\).

By two angles, \(\triangle PAB\sim\triangle PDC\). From corresponding sides, \(\frac{PA}{PD}=\frac{AB}{CD}\). Hence \(PA=50\cdot\frac{8}{10}=40\).

Method comment. equal inscribed angles turn triangles \(PAB\) and \(PDC\) into similar triangles Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(P,A,C\) are collinear, \(\angle PAB=\angle CAB\). Since \(P,D,B\) are collinear, \(\angle PDC=\angle CDB\).

Angles \(\angle CAB\) and \(\angle CDB\) are inscribed angles subtending the same chord \(CB\), so they are equal. Hence \(\angle PAB=\angle PDC\).

Similarly, \(\angle PBA=\angle DBA\), \(\angle PCD=\angle DCA\), and the inscribed angles \(\angle DBA\) and \(\angle DCA\) subtend chord \(DA\). Therefore \(\angle PBA=\angle PCD\).

By two angles, \(\triangle PAB\sim\triangle PDC\). From corresponding sides, \(\frac{PA}{PD}=\frac{AB}{CD}\). Hence \(PA=26\cdot\frac{9}{13}=18\).

Method comment. equal inscribed angles turn triangles \(PAB\) and \(PDC\) into similar triangles Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Let \(l\) be the tangent to \(\Omega\) at \(A\). By the statement, \(BD\parallel l\).

By the tangent-chord theorem, the angle between \(l\) and \(AB\) equals \(\angle ACB\). Since \(BD\parallel l\), we get \(\angle ABD=\angle ACB\).

Also \(D\in AC\), so \(\angle BAD=\angle BAC\). Therefore \(\triangle ABD\sim\triangle ACB\) by two angles.

From the similarity, \(\frac{AB}{AC}=\frac{AD}{AB}\). Multiplying gives \(AB^2=AC\cdot AD\), as required.

Method comment. the line parallel to the tangent transfers the tangent-chord theorem inside the triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(P,A,C\) are collinear, \(\angle PAB=\angle CAB\). Since \(P,D,B\) are collinear, \(\angle PDC=\angle CDB\).

Angles \(\angle CAB\) and \(\angle CDB\) are inscribed angles subtending the same chord \(CB\), so they are equal. Hence \(\angle PAB=\angle PDC\).

Similarly, \(\angle PBA=\angle DBA\), \(\angle PCD=\angle DCA\), and the inscribed angles \(\angle DBA\) and \(\angle DCA\) subtend chord \(DA\). Therefore \(\angle PBA=\angle PCD\).

By two angles, \(\triangle PAB\sim\triangle PDC\). From corresponding sides, \(\frac{PA}{PD}=\frac{AB}{CD}\). Hence \(PA=40\cdot\frac{11}{10}=44\).

Method comment. equal inscribed angles turn triangles \(PAB\) and \(PDC\) into similar triangles Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. By the tangent-chord theorem, \(\angle TAB=\angle ACB\), and \(\angle TAC=\angle ABC\). These two replacements are the main step of the problem.

Apply the sine rule in triangles \(TAB\) and \(TAC\). Since the angles at \(T\) are supplementary as angles on line \(BC\), after dividing the corresponding equalities we obtain \(\frac{TB}{TC}=\frac{\sin^2\angle ACB}{\sin^2\angle ABC}\).

In triangle \(ABC\), by the sine rule, \(\frac{AB}{AC}=\frac{\sin\angle ACB}{\sin\angle ABC}\). Hence \(\frac{TB}{TC}=\frac{AB^2}{AC^2}=\frac{121}{144}\). Therefore \(TB:TC=121:144\) in directed lengths.

Method comment. the tangent replaces angles at \(T\) by angles of the original triangle, after which the sine rule works Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(B,D,P,M\) lie on one circle, \(\angle BMP=\angle BDP\): both angles subtend chord \(BP\).

Points \(B,D,A\) are collinear, and points \(D,P,C\) are collinear. Hence \(\angle BDP\) is the oriented angle between lines \(AB\) and \(CD\).

Similarly, from the cyclic quadrilateral \(C,E,P,M\), we get \(\angle PME=\angle PCE\). But \(P,C,D\) are collinear and \(C,E,A\) are collinear, so \(\angle PCE\) is the oriented angle between \(CD\) and \(AC\).

Adding these two angles, we see that \(\angle BME\) equals the angle between \(AB\) and \(AC\), that is \(\angle BAC=69^\circ\).

But \(\angle BAE=\angle BAC\), since \(E\in AC\). Therefore \(\angle BME=\angle BAE\). By the converse criterion for equal inscribed angles, \(A,B,E,M\) lie on one circle.

Method comment. the second intersection of the two circles is a Miquel point and must also lie on the circle through \(A,B,E\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. By the tangent-chord theorem, \(\angle TAB=\angle ACB\), and \(\angle TAC=\angle ABC\). These two replacements are the main step of the problem.

Apply the sine rule in triangles \(TAB\) and \(TAC\). Since the angles at \(T\) are supplementary as angles on line \(BC\), after dividing the corresponding equalities we obtain \(\frac{TB}{TC}=\frac{\sin^2\angle ACB}{\sin^2\angle ABC}\).

In triangle \(ABC\), by the sine rule, \(\frac{AB}{AC}=\frac{\sin\angle ACB}{\sin\angle ABC}\). Hence \(\frac{TB}{TC}=\frac{AB^2}{AC^2}=\frac{169}{225}\). Therefore \(TB:TC=169:225\) in directed lengths.

Method comment. the tangent replaces angles at \(T\) by angles of the original triangle, after which the sine rule works Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Let \(l\) be the tangent to \(\Omega\) at \(A\). By the statement, \(BD\parallel l\).

By the tangent-chord theorem, the angle between \(l\) and \(AB\) equals \(\angle ACB\). Since \(BD\parallel l\), we get \(\angle ABD=\angle ACB\).

Also \(D\in AC\), so \(\angle BAD=\angle BAC\). Therefore \(\triangle ABD\sim\triangle ACB\) by two angles.

From the similarity, \(\frac{AB}{AC}=\frac{AD}{AB}\). Multiplying gives \(AB^2=AC\cdot AD\), as required.

Method comment. the line parallel to the tangent transfers the tangent-chord theorem inside the triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(P,A,C\) are collinear, \(\angle PAB=\angle CAB\). Since \(P,D,B\) are collinear, \(\angle PDC=\angle CDB\).

Angles \(\angle CAB\) and \(\angle CDB\) are inscribed angles subtending the same chord \(CB\), so they are equal. Hence \(\angle PAB=\angle PDC\).

Similarly, \(\angle PBA=\angle DBA\), \(\angle PCD=\angle DCA\), and the inscribed angles \(\angle DBA\) and \(\angle DCA\) subtend chord \(DA\). Therefore \(\angle PBA=\angle PCD\).

By two angles, \(\triangle PAB\sim\triangle PDC\). From corresponding sides, \(\frac{PA}{PD}=\frac{AB}{CD}\). Hence \(PA=35\cdot\frac{8}{7}=40\).

Method comment. equal inscribed angles turn triangles \(PAB\) and \(PDC\) into similar triangles Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(B,D,P,M\) lie on one circle, \(\angle BMP=\angle BDP\): both angles subtend chord \(BP\).

Points \(B,D,A\) are collinear, and points \(D,P,C\) are collinear. Hence \(\angle BDP\) is the oriented angle between lines \(AB\) and \(CD\).

Similarly, from the cyclic quadrilateral \(C,E,P,M\), we get \(\angle PME=\angle PCE\). But \(P,C,D\) are collinear and \(C,E,A\) are collinear, so \(\angle PCE\) is the oriented angle between \(CD\) and \(AC\).

Adding these two angles, we see that \(\angle BME\) equals the angle between \(AB\) and \(AC\), that is \(\angle BAC=54^\circ\).

But \(\angle BAE=\angle BAC\), since \(E\in AC\). Therefore \(\angle BME=\angle BAE\). By the converse criterion for equal inscribed angles, \(A,B,E,M\) lie on one circle.

Method comment. the second intersection of the two circles is a Miquel point and must also lie on the circle through \(A,B,E\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(B,D,P,M\) lie on one circle, \(\angle BMP=\angle BDP\): both angles subtend chord \(BP\).

Points \(B,D,A\) are collinear, and points \(D,P,C\) are collinear. Hence \(\angle BDP\) is the oriented angle between lines \(AB\) and \(CD\).

Similarly, from the cyclic quadrilateral \(C,E,P,M\), we get \(\angle PME=\angle PCE\). But \(P,C,D\) are collinear and \(C,E,A\) are collinear, so \(\angle PCE\) is the oriented angle between \(CD\) and \(AC\).

Adding these two angles, we see that \(\angle BME\) equals the angle between \(AB\) and \(AC\), that is \(\angle BAC=61^\circ\).

But \(\angle BAE=\angle BAC\), since \(E\in AC\). Therefore \(\angle BME=\angle BAE\). By the converse criterion for equal inscribed angles, \(A,B,E,M\) lie on one circle.

Method comment. the second intersection of the two circles is a Miquel point and must also lie on the circle through \(A,B,E\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(D\) is the foot of the altitude from \(B\), we have \(BD\perp AC\). Point \(H\) lies on this altitude, and \(A,D,C\) are collinear; hence \(DH\perp AD\), so \(\angle ADH=90^\circ\).

Similarly, \(E\) is the foot of the altitude from \(C\), so \(EH\perp AE\) and \(\angle AEH=90^\circ\).

Thus both \(D\) and \(E\) see segment \(AH\) under a right angle. Therefore \(D\) and \(E\) lie on the circle with diameter \(AH\). Hence \(A,D,H,E\) are concyclic.

In this circle, angles \(\angle DHE\) and \(\angle DAE\) subtend the same chord \(DE\). Therefore \(\angle DHE=\angle DAE=\angle BAC=61^\circ\).

Method comment. points \(D\) and \(E\) see segment \(AH\) under a right angle, so they should be placed on one circle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. By the tangent-chord theorem, \(\angle TAB=\angle ACB\), and \(\angle TAC=\angle ABC\). These two replacements are the main step of the problem.

Apply the sine rule in triangles \(TAB\) and \(TAC\). Since the angles at \(T\) are supplementary as angles on line \(BC\), after dividing the corresponding equalities we obtain \(\frac{TB}{TC}=\frac{\sin^2\angle ACB}{\sin^2\angle ABC}\).

In triangle \(ABC\), by the sine rule, \(\frac{AB}{AC}=\frac{\sin\angle ACB}{\sin\angle ABC}\). Hence \(\frac{TB}{TC}=\frac{AB^2}{AC^2}=\frac{100}{256}\). Therefore \(TB:TC=100:256\) in directed lengths.

Method comment. the tangent replaces angles at \(T\) by angles of the original triangle, after which the sine rule works Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. By the tangent-chord theorem, \(\angle TAB=\angle ACB\), and \(\angle TAC=\angle ABC\). These two replacements are the main step of the problem.

Apply the sine rule in triangles \(TAB\) and \(TAC\). Since the angles at \(T\) are supplementary as angles on line \(BC\), after dividing the corresponding equalities we obtain \(\frac{TB}{TC}=\frac{\sin^2\angle ACB}{\sin^2\angle ABC}\).

In triangle \(ABC\), by the sine rule, \(\frac{AB}{AC}=\frac{\sin\angle ACB}{\sin\angle ABC}\). Hence \(\frac{TB}{TC}=\frac{AB^2}{AC^2}=\frac{121}{324}\). Therefore \(TB:TC=121:324\) in directed lengths.

Method comment. the tangent replaces angles at \(T\) by angles of the original triangle, after which the sine rule works Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Let \(l\) be the tangent to \(\Omega\) at \(A\). By the statement, \(BD\parallel l\).

By the tangent-chord theorem, the angle between \(l\) and \(AB\) equals \(\angle ACB\). Since \(BD\parallel l\), we get \(\angle ABD=\angle ACB\).

Also \(D\in AC\), so \(\angle BAD=\angle BAC\). Therefore \(\triangle ABD\sim\triangle ACB\) by two angles.

From the similarity, \(\frac{AB}{AC}=\frac{AD}{AB}\). Multiplying gives \(AB^2=AC\cdot AD\), as required.

Method comment. the line parallel to the tangent transfers the tangent-chord theorem inside the triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(PX\perp BC\) and \(PY\perp CA\), angles \(\angle PXC\) and \(\angle PYC\) are right angles. Therefore \(P,X,C,Y\) lie on one circle.

Similarly, \(PZ\perp AB\) and \(PY\perp CA\), so \(\angle PZA=\angle PYA=90^\circ\), and points \(P,Z,A,Y\) also lie on one circle.

From the first circle, \(\angle XYC=\angle XPC\). From the second circle, \(\angle ZYA=\angle ZPA\).

Now compare the right-hand sides. Since \(PX\perp BC\), \(\angle XPC=90^\circ-\angle BCP\). Since \(PZ\perp AB\), \(\angle ZPA=90^\circ-\angle BAP\).

Points \(A,B,C,P\) lie on one circle, so \(\angle BCP=\angle BAP\): these are inscribed angles subtending chord \(BP\). Hence \(\angle XPC=\angle ZPA\), and therefore \(\angle XYC=\angle ZYA\).

Since \(A,Y,C\) are collinear, equality \(\angle XYC=\angle ZYA\) means that rays \(YX\) and \(YZ\) lie on one line. Thus \(X,Y,Z\) are collinear.

Method comment. two circles with diameters \(PC\) and \(PA\) give equal angles at point \(Y\) Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

E. Full solution. Since \(D\) is the foot of the altitude from \(B\), we have \(BD\perp AC\). Point \(H\) lies on this altitude, and \(A,D,C\) are collinear; hence \(DH\perp AD\), so \(\angle ADH=90^\circ\).

Similarly, \(E\) is the foot of the altitude from \(C\), so \(EH\perp AE\) and \(\angle AEH=90^\circ\).

Thus both \(D\) and \(E\) see segment \(AH\) under a right angle. Therefore \(D\) and \(E\) lie on the circle with diameter \(AH\). Hence \(A,D,H,E\) are concyclic.

In this circle, angles \(\angle DHE\) and \(\angle DAE\) subtend the same chord \(DE\). Therefore \(\angle DHE=\angle DAE=\angle BAC=55^\circ\).

Method comment. points \(D\) and \(E\) see segment \(AH\) under a right angle, so they should be placed on one circle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.

If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.

By the two-secant theorem, \(5\cdot18=6\cdot PD\). Hence \(PD=15\).

We have \(PT^2=3\cdot27=81\). Therefore \(PT=9\).

By the intersecting-chords theorem, \(XA\cdot XB=XC\cdot XD\). Hence \(8\cdot6=4\cdot XD\), so \(XD=12\).

By the tangent formula, the power of point \(P\) with respect to the circle equals \(PA^2\). The same power also equals \(PB^2\). Therefore \(PA^2=PB^2\), and since lengths are positive, \(PA=PB\).

Join \(A\) to \(C\), and \(B\) to \(D\). Consider triangles \(PAC\) and \(PDB\). The angle at \(P\) is common, because \(P,A,B\) and \(P,C,D\) are collinear. Also, \(\angle PAC=\angle BAC\), \(\angle PDB=\angle CDB\), and angles \(\angle BAC\) and \(\angle CDB\) are equal as inscribed angles standing on chord \(BC\). Therefore \(\triangle PAC\sim\triangle PDB\). From similarity, \(\frac{PA}{PD}=\frac{PC}{PB}\), hence \(PA\cdot PB=PC\cdot PD\).

Let the circle through \(A,B,C\) meet ray \(PC\) for the second time at \(D'\). By the secant theorem, \(PA\cdot PB=PC\cdot PD'\). By the condition, \(PA\cdot PB=PC\cdot PD\), hence \(PD'=PD\). Since \(D\) and \(D'\) lie on the same ray from \(P\), they coincide. Therefore \(A,B,C,D\) are cyclic.

Let \(XA=2k\), \(XB=5k\). Then \(XA\cdot XB=10k^2\). By the chord theorem, \(10k^2=XC\cdot XD=6\cdot15=90\). Hence \(k^2=9\), \(k=3\). Therefore \(XA=6\), \(XB=15\).

Let \(PA=x\). Then \(PB=x+20\). By the tangent-secant formula, \(12^2=x(x+20)\), so \(x^2+20x-144=0\). The positive root is \(x=6\). Hence \(PA=6\).

Lines \(AB\) and \(AC\) are secants of circle \((BCDE)\). Therefore \(AD\cdot AB=AE\cdot AC\). We get \(4\cdot14=7\cdot AC\), hence \(AC=8\).

For the first circle, the power of point \(P\) equals \(PA\cdot PB\). For the second circle, the intersection points with line \(PAB\) are the same \(A\) and \(B\), so the power also equals \(PA\cdot PB\). Hence the powers are equal.

By the tangent-secant formula, \(PT^2=PA\cdot PB\). Thus \(225=9\cdot PB\), so \(PB=25\). The secant passes through the centre, hence \(AB\) is a diameter. We get \(AB=PB-PA=25-9=16\).

By the secant theorem, \(PA\cdot PB=PC\cdot PD\). Hence \(\frac{PD}{PB}=\frac{PA}{PC}=\frac{2}{3}\). Therefore \(PD:PB=2:3\).

From point \(A\), two secants to circle \((BCDE)\) are drawn: \(ADB\) and \(AEC\). Therefore \(AD\cdot AB=AE\cdot AC\). Dividing both sides by \(AE\cdot AB\), we get \(\frac{AD}{AE}=\frac{AC}{AB}\).

On ray \(AB\) lie points \(D,B\), and on ray \(AC\) lie points \(E,C\). The equality \(AD\cdot AB=AE\cdot AC\) has the form of equality of products of two secants from point \(A\). By the converse power criterion, points \(D,B,E,C\), that is \(B,C,D,E\), lie on one circle.

Diagonals \(AC\) and \(BD\) are chords of one circle and meet at \(P\). By the intersecting-chords theorem, \(PA\cdot PC=PB\cdot PD\).

Line \(PAB\) meets the circle at \(A\) and \(B\), while line \(PCD\) meets it at \(C\) and \(D\). Therefore, by the two-secant theorem, \(PA\cdot PB=PC\cdot PD\).

By the tangent-secant theorem, \(TA^2=TB\cdot TC\). By the tangent-chord theorem, \(\angle TAB=\angle ACB\) and \(\angle TAC=\angle ABC\), so \(\triangle TAB\sim\triangle TCA\). From similarity, \(\frac{TB}{TA}=\frac{AB}{AC}\) and \(\frac{TA}{TC}=\frac{AB}{AC}\). Multiplying, we get \(\frac{TB}{TC}=\frac{AB^2}{AC^2}\).

With respect to circle \((ABD)\), from point \(C\) we have tangent \(CE\) and secant \(CDB\). Since \(D\) lies on side \(BC\), the points on the secant are in the order \(C,D,B\). By the tangent-secant formula, \(CE^2=CD\cdot CB\).

The power of point \(P\) with respect to the first circle equals \(PT_1^2\), and with respect to the second equals \(PT_2^2\). If \(PT_1=PT_2\), then \(PT_1^2=PT_2^2\). Therefore the powers are equal.

Draw the circle through \(A,B,C\). Let it meet line \(l_2\) for the second time at \(D'\), located on the same ray from \(P\) as \(D\). By the secant theorem, \(PA\cdot PB=PC\cdot PD'\). Comparing with the condition, we get \(PD'=PD\), hence \(D'=D\). Therefore \(A,B,C,D\) lie on one circle.

By the result proved above for a tangent to the circumcircle, \(\frac{TB}{TC}=\frac{AB^2}{AC^2}\). For completeness: it follows from the similarity \(\triangle TAB\sim\triangle TCA\), obtained by the tangent-chord theorem. For \(AB=9\), \(AC=6\), we get \(TB:TC=9^2:6^2=81:36=9:4\).

Equality of powers for point \(X\) gives \(XO_1^2-r_1^2=XO_2^2-r_2^2\), that is \(XO_1^2-XO_2^2=r_1^2-r_2^2\). The same equality holds for point \(Y\). Thus points \(X\) and \(Y\) have the same difference of squares of distances to \(O_1\) and \(O_2\). The locus of such points is a line perpendicular to \(O_1O_2\). Therefore \(XY\perp O_1O_2\).