Question 20
Let $ABC$ be a triangle with $2 \cdot \angle A \gt \angle B$. 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.)
(Purevsuren D.)