Problem

Source: Polish MO Finals P1 2024

Tags: geometry, geometry proposed, rectangle, angles



Let $X$ be an interior point of a rectangle $ABCD$. Let the bisectors of $\angle DAX$ and $\angle CBX$ intersect in $P$. A point $Q$ satisfies $\angle QAP=\angle QBP=90^\circ$. Show that $PX=QX$.