如图,在RT三角形ABC中,AC等于BC,∠ACB等于90°,点P,Q在AB上,且∠PCQ等于45°,试想线段AP,BQ

∵∠BCP =∠BCQ + 45° ∠AQC = ∠BCQ + ∠B =∠BCQ + 45°（ 外角和）
∴∠BCP=∠AQC
∵△ACQ∽△BCP(∠B=∠A,∠BCP=∠AQC)
∴(PA+PQ)/AC=BC/(PQ+BQ)
(PA+PQ)/BC=BC/(PQ+BQ)
整理BC^2=PQ^2+AP*BQ+PQ*AP+PQ*BQ (1)
而AB^2=2BC^2
AB=AP+PQ+BQ
AB^2=(AP+PQ+BQ)^2
=AP^2+PQ^2+BQ^2+2AP*PQ+2AP*BQ+2PQ*BQ
再将1式代入
2BC^2=2PQ^2+2AP*BQ+2PQ*AP+2PQ*BQ
=AP^2+PQ^2+BQ^2+2AP*PQ+2AP*BQ+2PQ*BQ
所以PQ^2=AP^2+BQ^2
直角三角形