Quadrilaterals and midpoints

Let AC intersect BD at X For convex case : Since $AX \perp BX$ then $AP=PX$ . Similarly, $AS=XS$ so that $PS \perp AX$ and we see that $APXS$ forms a kite. The same could be said about $CQXR$ . Thus, we get $PS \parallel QR \parallel BD$ . Moreover, we have $PS = \frac{1}{2}BD = QR$ by homothety. Analogously, we get $PQ=RS$ . Finally, $PQ \perp QR$ since $BX \perp XA, PQ \perp BX, AX \perp PS$ . The same goes for the other 3 points. For concave case, the same argument as above can be used. However, to prove that $PS \parallel QR \parallel BD$ we consider the homothety that maps QR to BD and PS to BD respectively instead of constructing the kites. The following arguments are the same as above. Hence, we arrive at the conclusion. P.S. sorry if I glossed over any details. P.P.S. I am not sure if this problem is original so sorry again if I offended anyone!