Inscribed trapezoids problem Let $ABCD$ be inscribed trapezoid with $(AB) \parallel (CD)$ and let $P$ be the point where its diagonals meet.
The circumcircle of $\triangle APB$ meets line $(BC)$ (again) at $X$.
$Y$ is a point on $(AX)$ such that $(DY)\parallel (BC)$.
Prove that $\angle YDA = 2 \angle YCA $.

Any hints would be appreciated.

Some hints:


*

*Prove that $ADPY$ is cyclic.

*Prove that $PC=PY$.
