Question

三角形ABC中,角ABC=90度,AB=BC,动点P在AC上(点P异于AC),将三角形BAP绕B点顺时针旋转90度得到△

Answer 1

PA^2+PC^2=2PB^2.证明:⊿BAP≌ΔBCQ,则AP=CQ;BP=BQ;∠A=∠BCQ.故:(1)∠BCQ+∠PCB=∠A+∠PCB=90°,则CQ^2+PC^2=PQ^2;(2)又∠PBQ=90°,则⊿PBQ为等腰直角三角形,PQ^2=PB^2+BQ^2=2PB^2.所以:PA^2+PC^2=CQ^2+PC^2=PQ^2=2...