Question 97
Consider a convex quadrilateral $ABCD$ with
$$
AB = CB \quad \text{and} \quad \angle ABC + 2\angle CDA = \pi
$$
and let $E$ be the midpoint of $AC$. Show that $\angle CDE = \angle BDA$.
$$
AB = CB \quad \text{and} \quad \angle ABC + 2\angle CDA = \pi
$$
and let $E$ be the midpoint of $AC$. Show that $\angle CDE = \angle BDA$.