#5 Circles I: Basic Circle Geometry

Angle \(\angle AOB\) is central, and \(\angle ACB\) is inscribed. They stand on the same arc \(AB\). Therefore \(\angle ACB=\frac{1}{2}\cdot96^\circ=48^\circ\).

Angles \(\angle ABC\) and \(\angle ADC\) are inscribed and stand on the same chord \(AC\). With the given position, they look at the same arc \(AC\), so they are equal.

The central angle corresponding to diameter \(AB\) is \(180^\circ\). The inscribed angle \(\angle ACB\) standing on the same arc is \(90^\circ\). Therefore \(AC\perp BC\).

We have \(OA=OB=OC=OD\), since these are radii of one circle. By condition \(AB=CD\). Hence \(\triangle AOB\cong\triangle COD\) by SSS, so \(\angle AOB=\angle COD\).

By the tangent theorem, the radius drawn to the point of tangency is perpendicular to the tangent. Since \(A\) is the point of tangency, \(OA\perp l\).

Since \(ABCD\) is cyclic, \(\angle A+\angle C=180^\circ\). Therefore \(\angle C=180^\circ-68^\circ=112^\circ\).

Both angles are inscribed and stand on chord \(AC\). With the given position, they are equal. Hence \(\angle ADC=41^\circ\).

The angle between the tangent at \(A\) and chord \(AC\) equals the inscribed angle standing on chord \(AC\). That angle is \(\angle ABC\). Therefore \(\angle ABC=37^\circ\).

Equal chords \(AB\) and \(AC\) cut off equal arcs. Angle \(\angle ADB\) stands on chord \(AB\), and angle \(\angle ADC\) stands on chord \(AC\). Therefore these inscribed angles are equal.

We have \(OA=OB\), since these are radii, \(AM=MB\) by condition, and \(OM\) is common. Hence \(\triangle OMA\cong\triangle OMB\). Therefore \(\angle OMA=\angle OMB\). These angles are supplementary, so each is \(90^\circ\). Thus \(OM\perp AB\).

If a point sees segment \(AB\) under a right angle, then it lies on the circle with diameter \(AB\). Since \(\angle ACB=90^\circ\) and \(\angle ADB=90^\circ\), both points \(C\) and \(D\) lie on this circle. Hence \(A,B,C,D\) lie on one circle.

The set of points from which segment \(AB\) is seen under a fixed angle lies on an arc of a circle through \(A\) and \(B\). Since \(C\) and \(D\) are on the same side of \(AB\) and \(\angle ACB=\angle ADB\), they lie on the same such arc. Therefore \(A,B,C,D\) lie on one circle.

The angle between the tangent and \(AB\) equals the inscribed angle standing on \(AB\), namely \(\angle ACB=43^\circ\). The angle between the tangent and \(AC\) equals the inscribed angle standing on \(AC\), namely \(\angle ABC=72^\circ\).

Since \(AB\parallel CD\), we have \(\angle ABD=\angle BDC\). In the cyclic quadrilateral, angles \(\angle BDC\) and \(\angle BAC\) stand on chord \(BC\), so they are equal. Hence \(\angle ABD=\angle BAC\). The first angle stands on chord \(AD\), and the second on chord \(BC\). Equal inscribed angles stand on equal chords, so \(AD=BC\).

Since \(BD\perp AC\) and \(D\in AC\), we get \(\angle BDC=90^\circ\). Since \(CE\perp AB\) and \(E\in AB\), we get \(\angle BEC=90^\circ\). Thus points \(D\) and \(E\) see segment \(BC\) under a right angle. Therefore they lie on the circle with diameter \(BC\), so \(B,C,D,E\) are cyclic.

By the tangent-chord theorem, the angle between the tangent and \(AB\) equals \(\angle ACB\). Since the tangent is parallel to \(BC\), the same angle equals \(\angle ABC\). Hence \(\angle ABC=\angle ACB\), so \(AB=AC\).

Equal chords \(AB\) and \(CD\) cut off equal arcs. Inscribed angles standing on equal arcs are equal. Therefore \(\angle ADB=\angle CAD\).

Since \(OM\perp AB\), point \(M\) is the midpoint of \(AB\). Similarly, \(N\) is the midpoint of \(CD\). From \(AB=CD\), we get \(AM=CN\). In right triangles \(OMA\) and \(ONC\), hypotenuses \(OA\) and \(OC\) are equal as radii, and legs \(AM\) and \(CN\) are equal. Thus the triangles are congruent, so \(OM=ON\).

Angle \(\angle BAC\) stands on chord \(BC\), and angle \(\angle CAD\) stands on chord \(CD\). By condition \(BC=CD\), so the corresponding arcs are equal. Therefore \(\angle BAC=\angle CAD\). Hence \(AC\) is the angle bisector of \(\angle BAD\).

Since \(OA\perp TA\) and \(OB\perp TB\), triangles \(OTA\) and \(OTB\) are right triangles. In them, \(OA=OB\) as radii, and \(OT\) is the common hypotenuse. Hence the triangles are congruent. Therefore \(TA=TB\), and also \(\angle ATO=\angle OTB\), so \(OT\) bisects angle \(ATB\).

Points \(A,D,B\) are collinear, so \(\angle BDE=180^\circ-\angle ADE\). By condition \(\angle ADE=\angle ACB\). Since \(E\) lies on \(AC\), angle \(\angle BCE=\angle BCA=\angle ACB\). Thus \(\angle BDE+\angle BCE=180^\circ\). The sum of opposite angles of quadrilateral \(BCDE\) is \(180^\circ\), therefore \(B,C,D,E\) lie on one circle.

Since \(P\) lies on \(AC\), angle \(\angle BAP=\angle BAC\). Since \(P\) lies on \(BD\), angle \(\angle ABP=\angle ABD\). Angle \(\angle APD\) is an exterior angle of triangle \(ABP\) at vertex \(P\). Therefore it equals the sum of the two remote interior angles: \(\angle APD=\angle BAP+\angle ABP=\angle BAC+\angle ABD\).

Since \(BD\perp AC\), we have \(\angle ADH=90^\circ\). Since \(CE\perp AB\), we have \(\angle AEH=90^\circ\). Thus points \(D\) and \(E\) see segment \(AH\) under a right angle, so \(A,D,H,E\) lie on the circle with diameter \(AH\). Also, \(\angle BDC=90^\circ\) and \(\angle BEC=90^\circ\), so \(D\) and \(E\) see segment \(BC\) under a right angle. Therefore \(B,C,D,E\) lie on the circle with diameter \(BC\).

Radii to the points of tangency are perpendicular to the tangents: \(OA\perp TA\) and \(OB\perp TB\). Therefore in quadrilateral \(AOBT\), the angles at \(A\) and \(B\) are \(90^\circ\). The sum of angles in a quadrilateral is \(360^\circ\), so \(\angle ATB=360^\circ-90^\circ-90^\circ-132^\circ=48^\circ\).

For the circle through \(A,B,C\), the tangent at \(B\) forms with chord \(BA\) an angle equal to the inscribed angle \(\angle BCA\), which stands on chord \(BA\). By condition, line \(l\) forms exactly this angle with \(BA\) and lies outside the triangle. Therefore \(l\) is the tangent to the circumcircle at point \(B\).

Angle \(\angle ABD\) stands on chord \(AD\), and angle \(\angle DBC\) stands on chord \(DC\). In one circle, equal inscribed angles stand on equal chords. Since \(\angle ABD=\angle DBC\), we get \(AD=DC\).

Since \(CT\parallel AB\), angle \(\angle ACT\) equals the angle between \(AC\) and \(AB\), that is \(\angle CAB\). By the tangent-chord theorem, \(\angle TAC=\angle ABC\). In triangle \(ACT\), equality \(AT=CT\) is equivalent to equality of base angles: \(\angle ACT=\angle TAC\). Hence \(AT=CT\) if and only if \(\angle CAB=\angle ABC\), that is, if and only if \(AC=BC\).

Since \(BD\perp AC\), line \(HD\) is perpendicular to \(AC\). Since \(CE\perp AB\), line \(HE\) is perpendicular to \(AB\). The angle between two lines equals the angle between their perpendiculars or supplements it to \(180^\circ\), depending on the chosen angle. Inside triangle \(DHE\), the angle is obtuse, so \(\angle DHE=180^\circ-\angle BAC=180^\circ-\angle A\).

Let \(O\) be the circumcentre. Radii \(OB\) and \(OC\) are perpendicular to tangents \(TB\) and \(TC\), so \(\angle OBT=\angle OCT=90^\circ\). The central angle \(\angle BOC\) standing on arc \(BC\) equals \(2\angle BAC\). In quadrilateral \(BOCT\), the angle sum is \(360^\circ\). Hence \(\angle BTC=360^\circ-90^\circ-90^\circ-\angle BOC=180^\circ-2\angle BAC\).

Since \(BD\perp AC\), we have \(\angle ADH=90^\circ\). Since \(CE\perp AB\), we have \(\angle AEH=90^\circ\). Thus \(A,D,H,E\) lie on the circle with diameter \(AH\). In this circle, angles \(\angle ADE\) and \(\angle AHE\) stand on the same chord \(AE\). Therefore \(\angle ADE=\angle AHE\).

E. Full solution.

Since \(PX\perp AB\) and \(PY\perp CB\), we have \(\angle PXB=\angle PYB=90^\circ\). Therefore points \(P,X,B,Y\) lie on one circle with diameter \(PB\).

From this cyclic quadrilateral, \(\angle XYB=\angle XPB\), because both angles subtend chord \(XB\). Thus the angle between \(XY\) and \(CB\) equals the angle between \(XP\) and tangent \(PB\).

Since \(XP\perp AB\), the angle between \(XP\) and \(PB\) equals \(90^\circ-\angle ABP\). By the tangent-chord theorem, \(\angle ABP=\angle ACB\).

Hence the angle between \(XY\) and \(CB\) is \(90^\circ-\angle ACB\). This exactly means that \(XY\perp AC\).

E. Full solution.

Assume the three initial points were not collinear. Then they lie on a unique circle \(\omega\).

We prove by induction that every newly marked point also lies on \(\omega\). Suppose chosen points \(A,B,C\) already lie on \(\omega\), and \(D\) is obtained by reflecting \(A\) across the perpendicular bisector of \(BC\).

The perpendicular bisector of chord \(BC\) passes through the center of circle \(\omega\). Hence \(\omega\) is symmetric with respect to this line. Therefore the reflection of point \(A\), which lies on \(\omega\), also lies on \(\omega\). Thus \(D\in\omega\).

So all marked points lie on one circle. But a line can meet a circle in at most two points, unless it coincides with the circle, which is impossible. Therefore three distinct marked points could not become collinear. This is a contradiction.

Hence, if three collinear points do appear, the three initial points must have been collinear.

E. Full solution.

Since \(PB\) and \(PC\) are tangents to \(\Omega\), radii \(OB\) and \(OC\) are perpendicular to them. Hence \(\angle OBP=\angle OCP=90^\circ\).

Thus points \(O,B,C,P\) lie on one circle, and \(OP\) is its diameter. Since \(S\) lies on this circle, \(\angle OSP=90^\circ\).

On the other hand, \(S\) lies on the circle with diameter \(AO\). Therefore \(\angle ASO=90^\circ\).

So both lines \(SP\) and \(SA\) pass through point \(S\) and are perpendicular to \(OS\). Through a point there is only one perpendicular to a given line; hence \(A,S,P\) are collinear.

E. Full solution.

On the extension of \(BC\) beyond \(C\), mark point \(N\) so that \(CN=CL\). Then \(BN=BC+CN=BC+CL\), so it is enough to prove \(BN=AB\).

Since \(A,B,K,L\) lie on one circle, angles \(\angle AKB\) and \(\angle ALB\) are supplementary. Using the collinearities \(A,K,C\) and \(B,L,C,N\), we get \(\angle CKB=\angle ALC\).

Triangles \(ACL\) and \(ACN\) are right triangles, have common leg \(AC\), and have equal legs \(CL=CN\). They are congruent, hence \(\angle ANC=\angle ALC\). Therefore \(\angle ANB=\angle CKB\).

Since \(BK\) bisects angle \(B\), \(\angle ABK=\angle KBC\). Using the right triangle and the previous angle equality, we obtain \(\angle BAN=\angle ANB\).

Thus triangle \(ABN\) is isosceles, and \(AB=BN\). Therefore \(AB=BC+CL\).

E. Full solution.

Since \(BH\perp AC\) and \(K\) lies on ray \(HB\), we have \(KH\perp AH\). Also, by construction, \(KM\perp AM\).

Thus \(\angle AHK=\angle AMK=90^\circ\), so points \(A,H,M,K\) lie on one circle. This circle has diameter \(AK\).

Angle \(\angle MAH\) equals \(\angle MAC=30^\circ\), because \(H\in AC\). Hence chord \(HM\), subtending a \(30^\circ\) angle, equals the radius of the circle, so \(HM=\frac{AK}{2}\).

In right triangle \(BHC\), point \(M\) is the midpoint of hypotenuse \(BC\), because \(AM\) is a median of triangle \(ABC\). Therefore \(HM=\frac{BC}{2}\).

We get \(\frac{AK}{2}=HM=\frac{BC}{2}\), hence \(AK=BC\).