14.3.2 角的平分线的判定-课件-2026-2027学年人教版数学八年级上册

#人教版八年级上册14.3.2角的平分线的判定专项练习（全文约1500字\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)可直接打印）##知识要点速记1.**角平分线的判定定理**：在角的内部\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)到角两边距离相等的点\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)在这个角的平分线上。符号语言：$\because$点$P$在$\angleAOB$内部\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)$PD\perpOA\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)PE\perpOB\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)\(PD=PE\)$$\therefore$点$P$在$\angleAOB$的平分线上（$OC$平分$\angleAOB$）⚠️关键点：①**必须在角的内部**；②必须是**垂线段相等**；两个条件缺一不可。2.和性质对比-性质（正向）：点在角平分线上$\boldsymbol{\Rightarrow}$到角两边距离相等-判定（逆向）：到角两边距离相等（角内）$\boldsymbol{\Rightarrow}$点在角平分线上3.三角形内心：三角形三条内角平分线交于一点（内心）\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)内心到三边距离相等；反过来：三角形内部到三边距离相等的点\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)是三条角平分线交点。4.证明思路：连接顶点和该点\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)用HL或者AAS证明两个直角三角形全等\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)推出角相等。⚠️易错提醒①缺少“垂直”\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)线段相等不能判定是角平分线；②点如果在角外部\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)即使到两边距离相等\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)也不在这个角的平分线上；③做题看清题目要求：让你证射线平分角\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)优先用判定定理。##一\(D\(BD、CE\)E\)基础过关（判断+填空+选择）###判断题（对√\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)错×）1.到角两边距离相等的点\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)一定在角平分线上。（）2.在角内部\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)到角两边垂线段相等的点\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)在角平分线上。（）3.$\(PA=PB\)$\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)就可以直接说明$OP$平分$\angleAOB$。（）4.三角形内部到三边距离相等的点\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)是三个内角平分线的交点。（）5.角平分线判定定理可以用AAS或者HL证明。（）###填空题6.在一个角的内部\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)到角两边的________相等的点\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)在这个角的平分线上。7.点$P$在$\angleAOB$内部\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)$PM\perpOA\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)PN\perpOB$\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)若$\(PM=PN\)$\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)则射线$OP$________$\angleAOB$。8.三角形内到三边距离相等的点叫做三角形的________。9.使用角平分线判定\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)两个必备条件：点在________；到两边的垂线段相等。10.角平分线性质：平分线$\Rightarrow$距离相等；角平分线判定：距离相等$\Rightarrow$________。###选择题11.下列条件\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)能判定$OP$平分$\angleAOB$的是（）A.$\(PA=PB\)$B.$PA\perpOA\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)PB\perpOB\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)\(PA=PB\)$C.$\angleA=\angleB$D.$\(OP=OP\)$12.三角形中\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)到三边距离相等的点是（）A.三条高线交点B.三条中线交点C.三条角平分线交点D.三边垂直平分线交点13.点$P$在$\angleABC$内部\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)$PD\perpAB\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)PE\perpBC$\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)$\(PD=PE\)$\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)则（）A.$\(PB=PD\)$B.$PB$平分$\angleABC$C.$\(PD=BC\)$D.$\(PE=AB\)$14.下面说法错误的是（）A.角平分线上的点到两边距离相等B.角外部到两边距离相等的点\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)不在这个角的平分线上C.只要两条线段相等\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)就能得到角平分线D.内心一定在三角形内部##二\(D\(BD、CE\)E\)中档解答题15.已知：点$P$在$\angleAOB$内部\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)$PD\perpOA$于$D$\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)$PE\perpOB$于$E$\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)$\(PD=PE\)$。求证：$OP$平分$\angleAOB$。16.已知：在$\triangleABC$中\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)$DE\perpAB$于$E$\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)$DF\perpAC$于$F$\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)$\(DE=DF\)$。求证：$AD$平分$\angleBAC$。17.已知：$BD\perpAC\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)CE\perpAB$\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)垂足为$D\(D\(BD、CE\)E\)E$\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)$\(BD=CE\)$\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)$BD\(D\(BD、CE\)E\)CE$交于点$P$。求证：$AP$平分$\angleBAC$。18.已知：$PM\perpAD\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)PN\perpCD$\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)$\(PM=PN\)$。求证：$DP$平分$\angleADC$。##三\(D\(BD、CE\)E\)拓展提升题19.已知：在四边形$ABCD$中\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)$\(AB=BC\)$\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)$BA\perpAD\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)BC\perpCD$。求证：$BD$平分$\angleADC$。20.已知：$\triangleABC$中\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)$\angleABC$和$\angleACB$外角的平分线交于点$P$\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)$PM\perpAB$\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)$PN\perpBC$\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)$PQ\perpAC$。求证：$\(PM=PQ\)$。##参考答案1.×（缺少“角内部+垂直”）2.√3.×（没有垂直）4.√5.√6.距离（垂线段长）7.平分8.内心9.角的内部10.点在角平分线上11.B12.C13.B14.C15.证明：连接$OP$$\becausePD\perpOA\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)PE\perpOB$$\therefore\angleODP=\angleOEP=90^\circ$在$Rt\triangleODP$和$Rt\triangleOEP$中$\begin{cases}\(OP=OP\)\\\(PD=PE\)\end{cases}$$\thereforeRt\triangleODP\congRt\triangleOEP(\text{HL})$$\therefore\angleDOP=\angleEOP$$\thereforeOP$平分$\angleAOB$16.证明：$\becauseDE\perpAB\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)DF\perpAC\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)\(DE=DF\)$\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)点$D$在$\angleBAC$内部$\thereforeAD$平分$\angleBAC$（角平分线的判定）17.证明：$\becauseBD\perpAC\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)CE\perpAB$$\therefore\anglePEA=\anglePDA=90^\circ$$\because\angleBEP=\angleCDP=90^\circ\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)\angleBPE=\angleCPD\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)\(BD=CE\)$可得$\(PE=PD\)$又$\becausePE\perpAB\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)PD\perpAC$$\thereforeAP$平分$\angleBAC$（角平分线判定）18.证明：$\becausePM\perpAD\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)PN\perpCD\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)\(PM=PN\)$\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)点$P$在$\angleADC$内部$\thereforeDP$平分$\angleADC$（角平分线的判定）19.证明：$\becauseBA\perpAD\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)BC\perpCD$$\therefore\angleA=\angleC=90^\circ$在$Rt\triangleABD$和$Rt\triangleCBD$中$\begin{cases}\(AB=BC\)\\BD=BD\end{cases}$$\thereforeRt\triangleABD\congRt\triangleCBD(\text{HL})$$\thereforeAD=CD$又$BA\perpAD\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)BC\perpCD$\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)点$B$在$\angleADC$内部$\thereforeBD$平分$\angleADC$20.证明：$\becauseBP$平分$\angleABC$的外角\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)$PM\perpAB\(PM\perpOA\(PD\perpAB\(BD\perpAC\(PM\perpAD\(BA\perpAD\(\becausePD\perpOA\(\becauseBD\perpAC\(\becausePE\perpAB\(\becauseBA\perpAD\(BA\perpAD\(PM\perpAB\(PN\perpBC，PQ\perpAC\)PN\perpBC\)BC\perpCD\)BC\perpCD\)PD\perpAC\)CE\perpAB\)PE\perpOB\)BC\perpCD\)PN\perpCD\)CE\perpAB\)PE\perpBC\)PN\perpOB\)PN\perpBC$$\therefore\(PM=PN\)$（角平分线性质）$\becauseCP$平分$\angleACB$的外角\(PM\perpOA