$\triangle PQR$ is an isosceles triangle where $PQ = PR$. Point $X$ is on the cicumcircle of $\triangle PQR$ such that it is in the opposite region of $P$ with respect to $QR$. $PY$ $\perp$ $XR$ and $XY$ = $12$. What is the value of $QX + XR$?
I am unable to solve the problem because I couldn't use the condition of $XY$ = $12$ and $\angle PYX$ = $90^\circ$ $\triangle PQR$ being an isosceles triangle. Then how can I suppose to get the value of $QX + XR$?
A small hint will be enough for me to proceed.
Source: Bangladesh Math Olympiad