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.
Best Answer
Some hints:
Prove that $ADPY$ is cyclic.
Prove that $PC=PY$.