Question

如图14，△ABC是等腰直角三角形，BC是斜边，P为△ABC内一点，将△ABP绕点A逆时针旋转后与△ACP 重合，如果AP=3，那么线段P P的长是多少？                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                             

 

Answer 1

 解：根据旋转的性质可知将△ABP绕点A逆时针旋转后与

  △ACP 重合△ABP≌△ACP,所以AP=A P,∠BAC=∠PA P=90°.所以在Rt△AP P

  中，P P=.