Prove that $AY$ is perpendicular to $BP$ > $ABC$ is a rt. angled triangle rt. angled at $A$. $ACPQ, BCYX, ABSR$ are squares drawn on $AC, BC , AB$ respectively. Prove that $AY$ is perpendicular to $BP$. ![enter image description here]( I couldnt proceed beyond congruency of the triangles $CPB$ and $CAY$. Please help.

After using rotation of $\Delta BCP$ on $90^{\circ}$ around the point $C$ we'll get $\Delta YCA$ we'll get $BP=AY$ and even $BP\perp AY$.