#7 Basic Constructions and Auxiliary Lines

The diagonals of quadrilateral \(ABDC\) are \(AD\) and \(BC\). By construction \(AM=MD\), and by condition \(BM=MC\). Hence the diagonals are bisected by point \(M\). Therefore \(ABDC\) is a parallelogram.

Since \(DE\parallel BC\), triangles \(ADE\) and \(ABC\) are similar. Since \(D\) is the midpoint of \(AB\), \(AD:AB=1:2\). Hence \(AE:AC=1:2\), so \(AE=EC\). Therefore \(E\) is the midpoint of \(AC\).

The set of points from which segment \(AB\) is seen under a right angle is the circle with diameter \(AB\). Since \(\angle ACB=90^\circ\), point \(C\) lies on this circle.

By construction, \(BD\parallel AC\) and \(CD\parallel AB\). Therefore in quadrilateral \(ABDC\), both pairs of opposite sides are parallel. Hence \(ABDC\) is a parallelogram.

Since \(M\) is the midpoint of \(BC\), \(BM=MC\). Since \(D\) is the reflection of \(A\) about \(M\), \(AM=MD\). Thus point \(M\) bisects diagonals \(BC\) and \(AD\) of quadrilateral \(ABDC\).

Since \(D\) lies on ray \(BA\), \(\angle DBC=\angle ABC=52^\circ\). In triangle \(BCD\), sides \(BD\) and \(BC\) are equal, so \(\angle BCD=\angle CDB\). Hence \(\angle BCD=\frac{180^\circ-52^\circ}{2}=64^\circ\).

Since \(DE\parallel AB\), triangles \(CDE\) and \(CBA\) are similar. We have \(CD:CB=3:5\), hence \(CE:CA=3:5\). Therefore \(CE:EA=3:2\).

Draw the circle with diameter \(AB\). Since \(\angle ACB=90^\circ\), point \(C\) lies on this circle. Since \(\angle ADB=90^\circ\), point \(D\) also lies on this circle. Therefore \(A,B,C,D\) lie on one circle.

By construction, \(ABDC\) is a parallelogram. In a parallelogram, the diagonals bisect each other. Here the diagonals are \(AD\) and \(BC\). Therefore the intersection of \(AD\) and \(BC\) is the midpoint of \(BC\).

Draw the circle through \(A,B,C\). On one side of \(AB\), all points from which segment \(AB\) is seen under angle \(\angle ACB\) lie on one arc of this circle. Since \(\angle ADB=\angle ACB\), point \(D\) lies on this circle. Hence \(A,B,C,D\) are cyclic.

Construct point \(P\) on the extension of \(MN\) so that \(NP=MN\). In triangles \(AMN\) and \(CNP\), we have \(AN=NC\), \(MN=NP\), and the angles at \(N\) are vertical. Hence the triangles are congruent, so \(AM=CP\) and \(AM\parallel CP\). Since \(AM=MB\), we get \(CP=MB\) and \(CP\parallel MB\), so \(BMCP\) is a parallelogram. Therefore \(MP\parallel BC\), hence \(MN\parallel BC\).

We have \(AM=MD\) by construction, \(BM=MC\) because \(M\) is the midpoint of \(BC\), and \(\angle AMB=\angle DMC\) as vertical angles. Therefore \(\triangle ABM\cong\triangle DCM\) by two sides and the included angle.

As in the basic construction, \(ABDC\) is a parallelogram. Therefore \(BD=AC\) and \(CD=AB\). By condition \(AB=AC\), hence \(BD=CD\).

Since \(DE\parallel AC\), triangles \(BDE\) and \(BAC\) are similar. From \(BD:BA=1:2\), it follows that \(BE:BC=1:2\). Hence \(BE=EC\), so \(E\) is the midpoint of \(BC\).

Since \(E\) lies on \(AC\), side \(AE\) is parallel to the line through \(D\) parallel to \(AC\), namely \(DF\). Since \(F\) lies on \(AB\), side \(AF\) is parallel to the line through \(D\) parallel to \(AB\), namely \(DE\). Thus \(AE\parallel DF\) and \(AF\parallel DE\), so \(AEDF\) is a parallelogram.

Since \(BD\perp AC\), and \(D\in AC\), we get \(\angle BDC=90^\circ\). Since \(CE\perp AB\), and \(E\in AB\), we get \(\angle BEC=90^\circ\). Hence points \(D\) and \(E\) lie on the circle with diameter \(BC\). Therefore \(B,C,D,E\) lie on one circle.

We have \(AM=MD\), \(BM=MC\), and \(\angle AMB=\angle DMC\) as vertical angles. Thus \(\triangle ABM\cong\triangle DCM\). Therefore \(\angle ABM=\angle DCM\). These are alternate interior angles for lines \(AB\) and \(CD\) with transversal \(BC\), so \(AB\parallel CD\).

Opposite sides of a parallelogram are equal. Therefore in \(ABDC\), \(AB=CD\) and \(AC=BD\). This is why a completed parallelogram often lets us move a segment to a convenient place.

Angles \(\angle ABC\) and \(\angle ADC\) are inscribed and stand on the same chord \(AC\). Therefore they are equal. In similar problems, a circle is useful because it lets us replace one angle by another angle located in a more convenient part of the diagram.

Since \(DE\parallel AC\), \(\triangle BDE\sim\triangle BCA\). From \(BD:DC=1:2\), we get \(BD:BC=1:3\). Therefore the areas of the similar triangles are in the square of the similarity ratio: \(S_{BDE}:S_{ABC}=1:9\).

Draw \(AC\). In triangle \(ABC\), segment \(MN\) is a midline, so \(MN\parallel AC\). In triangle \(CDA\), segment \(PQ\) is a midline, so \(PQ\parallel AC\). Therefore \(MN\parallel PQ\). Similarly, drawing \(BD\), we get \(NP\parallel MQ\). Hence \(MNPQ\) is a parallelogram.

Draw \(AC\), and let \(P\) be its midpoint. In triangle \(ABC\), segment \(MP\) is a midline, so \(MP\parallel BC\). In triangle \(ACD\), segment \(PN\) is a midline, so \(PN\parallel AD\). Since \(AD\parallel BC\), lines \(MP\) and \(PN\) coincide. Thus \(M,P,N\) are collinear, and \(MN\parallel AD\).

Triangles \(ABM\) and \(ACM\) are right triangles. They have \(BM=CM\), since \(M\) is the midpoint, and \(AM\) is common. Thus the triangles are congruent by two legs, so \(AB=AC\). The natural action here is to look for congruent triangles around the median.

Since \(M\) is the midpoint of \(BC\), we have \(MB=MC\). By condition \(AM=BM\). Hence \(MA=MB=MC\), and points \(A,B,C\) lie on the circle with centre \(M\). Also, \(BC\) is a diameter of this circle. Angle \(\angle BAC\) stands on the diameter, so \(\angle BAC=90^\circ\).

Draw \(AC\). Since \(AB\parallel CD\), we have \(\angle BAC=\angle ACD\). Also \(AB=CD\), and \(AC\) is common. Therefore \(\triangle BAC\cong\triangle DCA\) by two sides and the included angle. Hence \(\angle BCA=\angle CAD\), so \(BC\parallel AD\). Thus both pairs of opposite sides are parallel, and \(ABCD\) is a parallelogram.

The natural construction is to take point \(D\) on the extension of \(AM\) beyond \(M\) so that \(MD=AM\). Then \(AM=MD\), \(BM=MC\), and \(\angle AMB=\angle DMC\) as vertical angles. Therefore \(\triangle ABM\cong\triangle DCM\). Similarly, \(\triangle ACM\cong\triangle DBM\). Thus the construction creates two pairs of congruent triangles and parallelogram \(ABDC\).