In triangle ABC, AC = BC and angle C = 90. Points P and Q are on AB} such that QCP = 45. Show that AP^2 + BQ^2 = PQ^2.
Hint: Rotate the diagram counter-clockwise around C by 90 degrees. Let P go to point P'. What is angle P'CQ?
+605 $AP=BQ\implies AP=PQ\sqrt{2}$ is desired. Follows trivially from LoC and coordinates as $\angle ACP=\angle BCP=45^{\circ}$. Can't think of anything synthetic...
