Question 103
In isosceles trapezoid $ABCD$ where $BC = AD$ and $AB \parallel CD$, point $P$ is the intersection of diagonals $AC$ and $BD$. Let $X$ be the second intersection point of $BC$ and circumcircle of triangle $APB$ and let $Y$ be a point on $AX$ such that $DY \parallel BC$ ($C$ and $Y$ are on different sides of line $AD$). Prove that $\angle YDA = 2 \times \angle YCA$.