《2025年点石成金金牌夺冠八年级数学上册人教版辽宁专版》

20.(8分)如图，在$\bigtriangleup ABC$中，点$D$在边$BC$上，$\angle BAD = 100^{\circ}$，$\angle ABC$的平分线交$AC$于点$E$，过点$E$作$EF\bot BA$交$BA$的延长线于点$F$，且$\angle AEF = 50^{\circ}$，连接$DE$.
(1)求$\angle CAD$的度数.
(2)求证：$DE$平分$\angle ADC$.
(3)若$AB = 7$，$AD = 4$，$CD = 8$，且$S_{\bigtriangleup ACD} = 15$，求$\bigtriangleup ABE$的面积.

答案：
20.
(1)解：$\because EF\perp AB,\therefore \angle F=90^{\circ}$. $\because \angle AEF=50^{\circ},\therefore \angle FAE=90^{\circ}-\angle AEF=40^{\circ}$. $\because \angle BAD=100^{\circ}$， $\therefore \angle CAD=180^{\circ}-\angle BAD-\angle FAE=40^{\circ}$.
(2)证明：如图，过点$E$分别作$EG\perp AD$于点$G$，$EH\perp BC$于点$H$. 由
(1)得，$\angle FAE=\angle CAD=40^{\circ},EF\perp BF$， $\therefore AE$平分$\angle FAG.\therefore EF=EG$. $\because BE$平分$\angle ABC,EH\perp BC,\therefore EF=EH$. $\therefore EG=EH.\therefore DE$平分$\angle ADC$.
(3)解：$\because S_{\triangle ACD}=S_{\triangle AED}+S_{\triangle ECD}=\frac{1}{2}AD· EG+\frac{1}{2}CD· EH=15$，$EG=EH$，$AD=4$，$CD=8$， $\therefore \frac{1}{2}×4× EG+\frac{1}{2}×8× EG=15.\therefore EG=\frac{5}{2}$. $\therefore EF=EG=\frac{5}{2}$ $\therefore S_{\triangle ABE}=\frac{1}{2}AB· EF=\frac{1}{2}×7×\frac{5}{2}=\frac{35}{4}$.