在Rt三角形ABC中,角ACB=90°,AC=BC,P、Q为斜边AB上两点,角PCQ=45°试说明AP²+BQ²=PQ²

因为△AQC∽△BCP(∠B=∠A,∠BCP=∠AQC)所以(PA+PQ)/BC=BC/(PQ+BQ)整理BC^2=PQ^2+AP*BQ+PQ*AP+PQ*BQ (1)而AB^2=2BC^2AB=AP+PQ+BQAB^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*...