Question 1153587
Perform the rotation.  Then angle PCP' = 90 degrees, so PP' = CP*sqrt(2).  Then PQ = PP'/2 and CQ = CP = CP', so angle B'PQ = angle BPQ = 45.

Since CA = BC, angle CBQ = 45 and angle CAP = 45.  Therefore, angle ACP + angle BCQ = 45.  But BQ = QP' and P'Q' = B'Q', so angle CQ'P' = angle C'Q'P', which means angle CQ'B = 90.

Finally, PQ = P'Q' = (AP + BQ)/2 = 1/2 and angle PCQ = 135, so angle CAP = 45 - angle Q'CB.  So by the Pythagorean Theorem on right triangle ACQ, AP^2 + BC^2 = PC^2, which means AP^2 + BQ^2 = PQ^2.