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.

Observation 1: $AYPD$ has a circumcircle. 
Proof: $\angle AYD = \angle AXB = \angle APD$.
Then $\angle PBC = \angle DAP =\angle DYP$, and $\angle PDY \angle PAY = \angle PBC$. 
So $\angle DYP = \angle PDY$, hence $PDY$ is an isosceles triangle: $PY=PD=PC$, so $PY=PC$. 
Thus $\angle ADY = \angle APY = \angle ACY + \angle PYC = 2\angle ACY$.
