问题发现:如图1,∠AOB=90°,OC平分∠AOB,把三角尺的直角顶点落在OC的任意一点P上,并使三角尺的两条直角边分别与OA、OB相交于点E、F.探究发现PE=PF(可以这样想:作PM
![](https://math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmo%3E%E2%8A%A5%3C%2Fmo%3E%3C%2Fmath%3E)
OA于点M,PN
![](https://math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmo%3E%E2%8A%A5%3C%2Fmo%3E%3C%2Fmath%3E)
OB于点N,易得PM=PN,∠PME=∠PNF=90°,∠MPE=∠NPF=90°-∠EPN,所以△PNM
![](https://math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmtext%3E%E2%89%8C%3C%2Fmtext%3E%3C%2Fmath%3E)
△PNF,所以PE=PF)
变式拓展:如图2,已知∠AOB=120°,OC平分∠AOB,P是OC上一点,∠EPF=60°,PE边与OA边相交于点E,PF边与射线OB的反向延长线相交于点F.
![](http://tikupic.21cnjy.com/2023/05/02/c8/a1/c8a162d8d390cbe5dbe499d6d885114c_m_424x209.png)