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⟹AP=PQ√2 is desired. Follows trivially from LoC and coordinates as ∠ACP=∠BCP=45∘. Can't think of anything synthetic...
