Question 159
Let $ABCD$ be an isosceles trapezoid with $AB \parallel CD$ and $\overline{BC} = \overline{AD}$. The parallel to $AD$ through $B$ meets the perpendicular to $AD$ through $D$ at the point $X$. Furthermore, the parallel to $BD$ drawn through $A$ meets the perpendicular to $BD$ through $D$ at the point $Y$. Prove that the points $C, X, D$ and $Y$ lie on a common circle.