如圖14,△ABC是等腰直角三角形,BC是斜邊,P為△ABC內(nèi)一點,將△ABP繞點A逆時針旋轉后與△ACP 重合,如果AP=3,那么線段P P的長是多少?                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                             

 

 解:根據(jù)旋轉的性質可知將△ABP繞點A逆時針旋轉后與

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

  中,P P=.

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