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Mongolian National Mathematical Olympiad

Geometry Difficulty 7.0 National Olympiad, round 2 Prove it Mongolia

Let $ABC$ be a triangle with $2 \cdot \angle A > \angle B$ . A point $D$ is chosen on the segment $BC$ so that $2 \cdot \angle CAD = \angle ABC$ . Let $H$ be the orthocenter of the triangle $ABC$ and $AD$ intersects the circumcircle of the triangle $AHC$ at $K$ , differently from $A$ . $AB$ intersects the circumcircle of the triangle $BDK$ at $E$ , differently from $B$ . Prove that $AE = CD$ .
(Purevsuren D.)

Solution

Let $\beta := \angle B$ . Since $AHKC$ is a cyclic quadrilateral, we have $\angle AKC = \angle AHC = 180^\circ - \beta$ . By assumption $\angle KAC = \frac{\beta}{2}$ . Hence $\angle ACK = 180^\circ - \angle AKC - \angle KAC = \frac{\beta}{2}$ . Therefore $\triangle AKC$ is an isosceles triangle.

Since $EBDK$ is a cyclic quadrilateral, $\angle EKD = 180^\circ - \beta$ . Hence $\angle AKE = 180^\circ - \angle EKD = \beta$ and $\angle AKE + \angle AKC = 180^\circ$ . This shows that the points $E$ , $K$ and $C$ are collinear.

By the law of sines for the triangle $ADC$ we get that
$\frac{AC}{\sin \angle ADC} = \frac{DC}{\sin \frac{\beta}{2}}. \qquad (1)$
Since $EBDK$ is cyclic, $AEC = 180^\circ - \angle ADC$ . By the law of sines for the triangle $AEC$ we get that
$\frac{AE}{\sin \frac{\beta}{2}} = \frac{AC}{\sin \angle AEC} = \frac{AC}{\sin \angle ADC}. \qquad (2)$
From (1) and (2), we see that $AE = CD$ .

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