Question 167
Let $ABC$ be a triangle with a right angle at $C$. Let $D$ be the foot of the altitude from $C$, and $Z$ the point of $[AB]$ such that $AC = AZ$. The bisector of $\widehat{BAC}$ meets $(CB)$ and $(CZ)$ at $X$ and $Y$ respectively. Show that the four points $B$, $X$, $Y$, $D$ lie on a common circle.