Question 18
In triangle $ABC$, let $D$ be the foot of the perpendicular from $A$ to line $BC$. Point $K$ lies inside triangle $ABC$ such that $\angle KAB = \angle KCA$ and $\angle KAC = \angle KBA$. The line through $K$ perpendicular to line $DK$ meets the circle with diameter $BC$ at points $X$ and $Y$. Prove that $AX \cdot DY = DX \cdot AY$.