《2025年名校课堂九年级数学上册北师大版》

10. (教材 P85 习题 T4 变式)如图,在$\triangle ABC$中,点$D,E,F分别在边AB,AC,BC$上,$DE// BC,EF// AB$,且$AD:DB= 3:5$,那么$CF:CB= $(A)

A. $5:8$
B. $3:8$
C. $3:5$
D. $2:5$

答案： A

11. 如图,在$\triangle ABC$中,$EF// CD,DE// BC$.
(1)求证：$AF:FD= AD:DB$.
(2)若$AB= 15,AD:BD= 2:1$,求$DF$的长.

(1) 证明：$\because EF// CD$，$\therefore \frac{AF}{FD}=$$\frac{AE}{EC}$。$\because DE// BC$，$\therefore \frac{AD}{DB}=$$\frac{AE}{EC}$。$\therefore \frac{AF}{FD}=\frac{AD}{DB}$，即$AF:FD = AD:DB$。
(2) $\because AD:BD = 2:1$，$\therefore$ 设$AD =$$2x$，$BD =$$x$。$\because AB = AD + BD$，$\therefore 15 =$$2x + x$，解得$x =$$5$。$\therefore AD =$$10$。由(1)，得$\frac{AF}{FD}=\frac{AD}{DB}=$$\frac{2}{1}$，$\therefore$ 设$AF =$$2y$，$FD =$$y$。$\because AD = AF + DF$，$\therefore 10 =$$2y + y$，解得$y =$$\frac{10}{3}$。$\therefore DF =$$\frac{10}{3}$。

答案： 解：
(1) 证明：$\because EF// CD$，$\therefore \frac{AF}{FD}=\frac{AE}{EC}$。$\because DE// BC$，$\therefore \frac{AD}{DB}=\frac{AE}{EC}$。$\therefore \frac{AF}{FD}=\frac{AD}{DB}$，即$AF:FD = AD:DB$。
(2) $\because AD:BD = 2:1$，$\therefore$ 设$AD = 2x$，$BD = x$。$\because AB = AD + BD$，$\therefore 15 = 2x + x$，解得$x = 5$。$\therefore AD = 10$。由
(1)，得$\frac{AF}{FD}=\frac{AD}{DB}=\frac{2}{1}$，$\therefore$ 设$AF = 2y$，$FD = y$。$\because AD = AF + DF$，$\therefore 10 = 2y + y$，解得$y = \frac{10}{3}$。$\therefore DF = \frac{10}{3}$。

12. 新考向 阅读理解 请阅读以下材料,并回答相应的问题：
角平分线分线段成比例定理：三角形内角平分线分对边所得的两条线段和这个角的两边对应成比例.
如图 1,在$\triangle ABC$中,$AD平分\angle BAC$,则$\frac {AB}{AC}= \frac {BD}{CD}$.
下面是这个定理的部分证明过程.
证明：如图 2,过点$C作CE// DA$,交$BA的延长线于点E$……
(1)请按照上面的证明思路,写出该证明的剩余部分.
(2)如图 3,在$Rt\triangle ABC$中,$AB= 3,BC= 4,\angle ABC= 90^{\circ },AD平分\angle BAC$,则$\triangle ABD$的周长是____.

(1)证明：$\because CE// AD$，$\therefore \frac{BD}{CD}=\frac{AB}{AE}$，$\angle 2 = \angle ACE$，$\angle 1 = \angle E$。$\because \angle 1 = \angle 2$，$\therefore \angle ACE = \angle E$。$\therefore AE = AC$。$\therefore \frac{AB}{AC}=\frac{BD}{CD}$。
(2)$\frac{3\sqrt{5}+9}{2}$

答案： 解：
(1) 证明：$\because CE// AD$，$\therefore \frac{BD}{CD}=\frac{AB}{AE}$，$\angle 2 = \angle ACE$，$\angle 1 = \angle E$。$\because \angle 1 = \angle 2$，$\therefore \angle ACE = \angle E$。$\therefore AE = AC$。$\therefore \frac{AB}{AC}=\frac{BD}{CD}$。【一题多解】过点$D$分别作$DF\perp AB$于点$F$，$DG\perp AC$于点$G$，$\because AD$平分$\angle BAC$，$\therefore DF = DG$。$\therefore \frac{S_{\triangle ABD}}{S_{\triangle ACD}}=\frac{\frac{1}{2}AB\cdot DF}{\frac{1}{2}AC\cdot DG}=\frac{AB}{AC}=\frac{BD}{CD}$。
(2) $\frac{3\sqrt{5}+9}{2}$

【例】 如图,在$\triangle ABC$中,$AF:FB= 2:3$,延长$BC至点D$,使得$BC= 2CD$,则$\frac {AE}{EC}$的值为2.
【提示】 过点$C作CG// DF交AB于点G$.

答案： 【例】2

【变式】 如图,$D,E分别是\triangle ABC边BC,AC$上的点,$BD:CD= 2:5$,连接$AD,BE$,交点为$F,DF:AF= 1:4$,那么$\frac {CE}{AE}$的值是____$\frac{7}{8}$.

答案： 【变式】$\frac{7}{8}$