Problem

Source: Iran MO 2024 3rd Round Geometry Exam P1

Tags: geometry



Let $ABCD$ be a cyclic quadrilateral with circumcircle $\Gamma$. Let $M$ be the midpoint of the arc $ABC$. The circle with center $M$ and radius $MA$ meets $AD, AB$ at $X, Y$. The point $Z \in XY$ with $Z \neq Y$ satisfies $BY=BZ$. Show that $\angle BZD=\angle BCD$.