For a quadrilateral $PQRS$, $PR=40$ and $PS=32$. Suppose for a point $X$ on the line segment $RS$ the triangles $PQX$ and $PRS$ are congruent in that order. If area of triangle $PXS$ is $112$ square units, find area of the triangle $QRX$.
I firstly noted that $PQRX$ is cyclic. Then used $\frac12ab\sin\theta$ for areas thrice and got some equations, which seem to be very messy to work with. And also I am one equation short (I took 2 sides and 3 angles as new variables).
Any help to proceed?