Question 96
On the side $BC$ of the acute triangle $ABC$ lies a point $D$ such that $|AB| = |AD|$. Let $E$ be such a point at the height from $C$ of the triangle $ABC$ that the circle $\mathcal{K}_1$ with its center in $E$ touches the line $AD$ at the point $D$. We denote by $\mathcal{K}_2$ the circle through $C$, which touches the line $AB$ at the point $B$. Prove that the point $A$ lies on the line passing through the intersections of the circles $\mathcal{K}_1$ and $\mathcal{K}_2$.