Question 29
Let $\overarc{ABCD}$ be a cyclic quadrilateral with incircle $\omega$. Let $M$ be the midpoint of arc $\overarc{ABC}$. Let $\Gamma$ be the circle with center $M$ and radius $MA$. $AD$ and $AB$ intersect $\Gamma$ at $X$ and $Y$, respectively. Let $Z$ be a point on line $XY$ (with $Z \neq Y$) such that $BY = BZ$. Prove that $\angle BZD = \angle BCD$.