Question 70
Let $ABCD$ be a convex quadrilateral such that $\angle ADB = \angle BDC$. Suppose that a point $E$ on the side $AD$ satisfies the equality
$$
AE \cdot ED + BE^2 = CD \cdot AE.
$$
Show that $\angle EBA = \angle DCB$.
$$
AE \cdot ED + BE^2 = CD \cdot AE.
$$
Show that $\angle EBA = \angle DCB$.