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AP+BQ=PQ

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^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 好好学习,天天向上.