A point $D$ is taken on the side $BC$ of an acute-angled triangle $ABC$ such that $AB = AD$. Point $E$ on the altitude from $C$ of the triangle is such that the circle $k_1$ with center $E$ is tangent to the line $AD$ at $D$. Let $k_2$ be the circle through $C$ that is tangent to $AB$ at $B$. Prove that $A$ lies on the line determined by the common chord of $k_1$ and $k_2$.