In triangle \( ABC, AC = BC\) and \( \angle ACB = 90^\circ.\)Points P and Q are on \(\overline AB\) such that P is between A and Q and \(\angle QCP = 45^\circ\). Show that \(AP^2 + BQ^2 = PQ^2\).
Hint(s): Rotate the diagram counter-clockwise around C by \(90^\circ\). Let P go to point P'. What is \(\angle P'CQ\)? Show your work.