5.4.2角平分线性质的应用 课件 -2026-2027学年湘教版数学八年级上册

#湘教版数学八年级上册5.\(\boldsymbol{4}\).2角平分线性质的应用专项练习##知识点回顾角平分线性质：角平分线上的点到角两边的垂线段距离相等。角平分线逆定理：在角的内部\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)到角两边距离相等的点\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)在这个角的平分线上。✅应用常见题型：1.求线段长度；2.证明线段相等；3.证明角相等；\(\boldsymbol{4}\).三角形内心（三条角平分线交点）\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)内心到三边距离相等；5.实际作图\(DB\(AM\(DB、DC\)BN\)DC\)选址问题（到角两边距离相等的点）。>核心条件：看到角平分线+垂直→直接得到垂线段相等。满分：100分时间：\(\boldsymbol{4}\)0分钟##一\(DB\(AM\(DB、DC\)BN\)DC\)选择题（每题\(\boldsymbol{4}\)分\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)共20分）1.如图\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$OP$平分$\angleAOB$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$PM\perpOA$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$PN\perpOB$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)若$\(PM=6\)$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)则$PN$的长为（）A.3B.6C.9D.122.在$\triangleABC$中\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$AD$是角平分线\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$DE\perpAB$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$DF\perpAC$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)下列一定成立的是（）A.$\(BD=CD\)$B.$\(DE=DF\)$C.$\(AB=AC\)$D.$\angleB=\angleC$3.三角形内一点到三边距离相等\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)这个点是三角形（）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\(\boldsymbol{4}\).$\triangleABC$中\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$\angleC=90^\circ$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$AD$平分$\angleCAB$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$\(BC=8\)$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$\(BD=5\)$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)则点$D$到$AB$的距离是（）A.3B.5C.8D.135.下列应用题\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)适合用角平分线性质解决的是（）A.求三角形内角和B.找区域内到两条道路距离相等的点C.求斜边长度D.证明两线段中点相同##二\(DB\(AM\(DB、DC\)BN\)DC\)填空题（每题\(\boldsymbol{4}\)分\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)共20分）1.角平分线上的点到角两边的________相等。2.$\triangleABC$中\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)角平分线交于点$I$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)则点$I$到________距离相等。3.$OC$平分$\angleAOB$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)点$P$在$OC$上\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$PM\perpOA$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$\(PM=\(\boldsymbol{7}\)\)$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)则$P$到$OB$距离为________。\(\boldsymbol{4}\).在直角三角形中\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)角平分线交点到直角边的距离\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)是利用________定理求解。5.要找一个点在角内部且到角两边距离相等\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)则该点一定在________。##三\(DB\(AM\(DB、DC\)BN\)DC\)基础解答题（每题8分\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)共32分）1.已知：在$\triangleABC$中\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$AD$平分$\angleBAC$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$DE\perpAB$于$E$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$DF\perpAC$于$F$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$\(DE=\(\boldsymbol{4}\)\)$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)求$DF$的长度。2.判断对错\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)并说明理由（1）$AD$平分$\angleBAC$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)点$D$在$AD$上\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)连接$DB\(DB\(AM\(DB、DC\)BN\)DC\)DC$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)则$\(DB=DC\)$。（2）三角形内心到三边的距离相等。3.已知：$\angleC=90^\circ$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$AD$平分$\angleCAB$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$DE\perpAB$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)垂足为$E$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$\(CD=3\)$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)求$DE$的长。\(\boldsymbol{4}\).如图\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$BD$平分$\angleABC$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$DE\perpAB$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$DF\perpBC$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$\(AB=8\)$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$\(BC=6\)$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$\(DE=5\)$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)求$DF$。##四\(DB\(AM\(DB、DC\)BN\)DC\)拓展提升题（每题1\(\boldsymbol{4}\)分\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)共28分）1.已知：在$\triangleABC$中\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$\angleC=90^\circ$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$AD$平分$\angleBAC$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$\(BC=10\)$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$\(BD=6\)$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)求点$D$到$AB$边的距离。2.已知：$\triangleABC$的角平分线$AM\(DB\(AM\(DB、DC\)BN\)DC\)BN$交于点$P$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$PD\perpAB$于$D$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$PE\perpBC$于$E$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$PF\perpAC$于$F$。求证：点$P$在$\angleC$的平分线上。---#参考答案与解析##一\(DB\(AM\(DB、DC\)BN\)DC\)选择题1.B2.B3.C\(\boldsymbol{4}\).A5.B##二\(DB\(AM\(DB、DC\)BN\)DC\)填空题1.**垂线段距离**2.**三边**3.$\boldsymbol{\(\boldsymbol{7}\)}$\(\boldsymbol{4}\).**角平分线**5.**这个角的平分线上**##三\(DB\(AM\(DB、DC\)BN\)DC\)基础解答题1.解：$\becauseAD$平分$\angleBAC\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)DE\perpAB\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)DF\perpAC$$\thereforeDF=\(DE=\(\boldsymbol{4}\)\)$（角平分线上的点到角两边距离相等）答：$DF=\boldsymbol{\(\boldsymbol{4}\)}$。2.（1）×；没有垂直条件\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$DB\(DB\(AM\(DB、DC\)BN\)DC\)DC$不是垂线段\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)不一定相等。（2）√；三角形三条角平分线交于内心\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)内心到三边距离相等。3.解：$\becauseAD$平分$\angleCAB\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)\angleC=90^\circ$即$DC\perpAC$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$DE\perpAB$$\thereforeDE=\(CD=3\)$答：$DE=\boldsymbol{3}$。\(\boldsymbol{4}\).解：$\becauseBD$平分$\angleABC\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)DE\perpAB\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)DF\perpBC$$\thereforeDF=\(DE=5\)$答：$DF=\boldsymbol{5}$。##四\(DB\(AM\(DB、DC\)BN\)DC\)拓展提升题1.解：$\because\(BC=10\)\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)\(BD=6\)$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$\thereforeCD=BC-BD=10-6=\(\boldsymbol{4}\)$$\becauseAD$平分$\angleBAC\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)\angleC=90^\circ\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)DE\perpAB$$\thereforeD$到$AB$距离$\(DE=CD=\(\boldsymbol{4}\)\)$答：点$D$到$AB$距离为$\boldsymbol{\(\boldsymbol{4}\)}$。2.证明：$\becauseAM$平分$\angleBAC\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)PD\perpAB\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)PF\perpAC$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$\thereforePD=PF$$\becauseBN$平分$\angleABC\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)PD\perpAB\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)PE\perpBC$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)$\thereforePD=PE$$\thereforePE=PF$又$\becausePE\perpBC\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)PF\perpAC$\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)点$P$在$\triangleABC$内部$\therefore$点$P$在$\angleC$的平分线上（角平分线逆定理）##易错警示1.做题前提**必须有两条垂直**\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)没有垂直不能直接用角平分线性质；2.区分：内心（角平分线交点\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)到三边距离相等）；外心（垂直平分线交点\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)到三顶点距离相等）；3.应用题审题：距离=垂线段\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)不是斜线段；\(\boldsymbol{4}\).逆定理使用\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)一定要说明点在角**内部**。第五章到此全部结束！下一章：**6.1.1函数**\(\angleCAB\(\becauseBC=10\(\becausePE\perpBC，PF\perpAC\)BD=6\)\angleC=90^\circ\)继续同模板出题吗？湘教版（新教材）数学八年级上册5.4.2角平分线性质的应用授课教师：.

班

级：

8年级（

）班.

时

间：.

2026/10/10

学子如同向日葵向往暖阳，满心热爱渴求知识。在书页间汲取力量，于思索中追逐光亮，不惧学习路上的磨砺，心怀憧憬奋力向上，向着理想不断成长。第5章直角三角形如图，在△ABC中，D，E，F分别是BC,AB，AC边上的点.若BE=CF，S∆BDE=S∆CDF，则点D在∠BAC的平分线上吗?新知探究说一说新知探究思

考如图，已知EF⊥CD于点E，EF⊥AB于点F，MN⊥AC于点N，M是EF的中点.需要添加一个什么条件，就可使CM，AM分别为∠ACD和∠CAB的平分线呢?CM为∠ACD的平分线分析：根据条件需MN=EMAM为∠CAB的平分线根据条件需MN=FM已知EM=FM

需要添加一个什么条件?MN=EMMN=FM新知探究添加条件MN=MF可以吗？若添加MN=EM因为ME⊥CD，MNLAC,MN=ME,所以点M在∠ACD的平分线上，即CM是∠ACD的平分线. 又M是EF的中点，则MF=ME=MN.同理可证AM是∠CAB的平分线.典例分析例2

如图，在△ABC的外角∠CAD的平分线上任取一点P，作PE⊥DB，PF⊥AC，垂足分别为点E，F.试探索BE+PF与PB的大小关系.分析：BE+PF必须把BE、PF转化到一条线段上或一个三角形中根据题意可以将PF转化成PEBE、PE、PB是一个三角形的三边典例分析解

因为AP是∠CAD的平分线，

又PE⊥DB，PF⊥AC，

所以PE=PF.在△EBP中，BE+PE>PB,因此BE+PF>PB.返回1．在正方形网格中，∠ACB的位置如图所示，则到∠ACB两边距离相等的点是(

)A．点MB．点NC．点PD．点QA

DA.

一处

B.

两处

C.

三处

D.

四处返回新知探究做一做任意作一个△ABC，在△ABC内部找一点P，使其到三边的距离相等.A

B

C

P

N

M

D

E

F

角的内部到角的两边距离相等的点在角的平分线上根据题意可知点P在∠ABC、∠ACB、∠BAC的平分线上找点P，只要画△ABC中任意两个内角的角平分线在△ABC中分别作∠BAC与∠ABC的平分线，它们交于点P这个点P为什么到三边的距离相等呢？新知探究A

B

C

P

N

M

D

E

F

证明.过点P作PD⊥AB,PE⊥AC,PF⊥BC，垂足分别为点D，E，F.因为AP是∠BAC的平分线，PD⊥AB，PE⊥AC，所以PD=PE.因为BP是∠ABC的平分线，PD⊥AB，PF⊥BC,所以PD=PF.故PD=PE=PF，因此P为所求作的点.结论：三角形的三条角平分线交于一点，并且这点到三边的距离相等.返回3．[泰安市期中]如图所示，点O在一块直角