Perpendicularity in a right-triangle

This problem was proposed by Anthony Becerra in the Facebook group "Romantics of Geometry".

Consider a right-triangle $ABC$ with $\angle{ABC}=90^\circ$. Let $P$, $Q$ and $R$ be on sides $AB$, $BC$ and $AC$, respectively, such that $BPRQ$ is a square. Call $X$ the intersection of $AQ$ and $CP$. Prove that $BX\perp{AB}$.

Proof. By Ceva's theorem, $\frac{AH}{CH}\cdot{\frac{CQ}{BQ}}\cdot{\frac{BP}{AP}}=1$. By similarity, $\frac{CR}{AC}=\frac{CQ}{BC}$; $\frac{BQ}{BC}=\frac{AR}{AC}$, then, $\frac{CQ}{BQ}=\frac{CR}{AR}$. But, because of the Angle Bisector theorem, $\frac{CR}{AR}=\frac{BC}{AB}$. Now, as $\triangle{APR}\sim\triangle{CQR}$, it follows $\frac{BP}{AP}=\frac{CR}{AR}=\frac{BC}{AB}$. Back to Ceva: $\frac{AH}{CH}\cdot{\frac{BC^2}{AB^2}}=1$, which implies the perpendicularity by the converse of Euclid's theorem.