《2025年名校课堂九年级数学上册北师大版》

9. 如图,在三角形纸片$ABC$中,$\angle A = 90^{\circ}$,$AB = 9$,$AC = 12$,折叠纸片,使点$A落在边BC上的点D$处,折痕为$EF$(点$E在AB$上,点$F在AC$上)。若$\triangle BDE与\triangle ABC$相似,则折痕$EF$的长度等于$\frac{36\sqrt{2}}{7}$或$4\sqrt{10}$。

答案： $\frac{36\sqrt{2}}{7}$或$4\sqrt{10}$

10. 如图$,\angle MAN = 30^{\circ},$点B,C分别在AM,AN上,且$\angle ABC = 40^{\circ}。$
(1)尺规作图：作$\angle CBM$的平分线BD,BD与AN相交于点D。(保留作图痕迹,不写作法)图略(2)在(1)所作的图中,求证：$\triangle ABC \backsim \triangle ADB。$证明$: \because \angle ABC = 40^{\circ}, \therefore \angle MBC = 140^{\circ}. \because BD$平分$\angle MBC, \therefore \angle MBD = \frac{1}{2}\angle MBC = 70^{\circ}. \because \angle ADB = \angle MBD - \angle A = 70^{\circ} - 30^{\circ} = 40^{\circ}, \therefore \angle ABC = \angle ADB. $又$\because \angle A = \angle A, \therefore \triangle ABC \backsim \triangle ADB.$

答案： 解:
(1) 图略.
(2) 证明: $\because \angle ABC = 40^{\circ}, \therefore \angle MBC = 140^{\circ}$. $\because BD$平分$\angle MBC, \therefore \angle MBD = \frac{1}{2}\angle MBC = 70^{\circ}$. $\because \angle ADB = \angle MBD - \angle A = 70^{\circ} - 30^{\circ} = 40^{\circ}, \therefore \angle ABC = \angle ADB$. 又$\because \angle A = \angle A, \therefore \triangle ABC \backsim \triangle ADB$.

11. 如图,已知$\triangle ABC \backsim \triangle AEF$,$B$,$E$,$F$三点共线,线段$EF与AC交于点O$。
(1)求证：$\triangle ABE \backsim \triangle ACF$。
(2)若$AF = 5$,$BC = 10$,$\triangle AOF的面积为8$,求$\triangle BOC$的面积。

(1) 证明: $\because \triangle ABC \backsim \triangle AEF, \therefore \angle BAC = \angle EAF, AB : AE = AC : AF$. $\therefore \angle BAC - \angle EAC = \angle EAF - \angle EAC$, 即$\angle BAE = \angle CAF, AB : AC = AE : AF$. $\therefore \triangle ABE \backsim \triangle ACF$.
(2) $\because \triangle ABC \backsim \triangle AEF, \therefore \angle ACB = \angle AFE$, 即$\angle AFO = \angle BCO$. 又$\because \angle AOF = \angle BOC, \therefore \triangle AOF \backsim \triangle BOC$. $\therefore S_{\triangle BOC} : S_{\triangle AOF} = (\frac{BC}{AF})^2 = 4$. $\because S_{\triangle AOF} = 8, \therefore S_{\triangle BOC} = $32.

答案： 解:
(1) 证明: $\because \triangle ABC \backsim \triangle AEF, \therefore \angle BAC = \angle EAF, AB : AE = AC : AF$. $\therefore \angle BAC - \angle EAC = \angle EAF - \angle EAC$, 即$\angle BAE = \angle CAF, AB : AC = AE : AF$. $\therefore \triangle ABE \backsim \triangle ACF$.
(2) $\because \triangle ABC \backsim \triangle AEF, \therefore \angle ACB = \angle AFE$, 即$\angle AFO = \angle BCO$. 又$\because \angle AOF = \angle BOC, \therefore \triangle AOF \backsim \triangle BOC$. $\therefore S_{\triangle BOC} : S_{\triangle AOF} = (\frac{BC}{AF})^2 = 4$. $\because S_{\triangle AOF} = 8, \therefore S_{\triangle BOC} = 32$.

12. 新考向 新定义问题 从三角形(不是等腰三角形)一个顶点引出一条射线与对边相交,顶点与交点之间的线段把这个三角形分割成两个小三角形,如果分得的两个小三角形中一个为等腰三角形,另一个与原三角形相似,我们把这条线段叫做这个三角形的完美分割线。
(1)如图1,在$\triangle ABC$中,$CD$为角平分线,$\angle A = 40^{\circ}$,$\angle B = 60^{\circ}$,求证：$CD为\triangle ABC$的完美分割线。
(2)如图2,在$\triangle ABC$中,$AC = 2$,$BC = \sqrt{2}$,$CD是\triangle ABC$的完美分割线,且$\triangle ACD是以CD$为底边的等腰三角形,求完美分割线$CD$的长。

(1) 证明: $\because \angle A = 40^{\circ}, \angle B = 60^{\circ}, \therefore \angle ACB = 80^{\circ}$. $\therefore \triangle ABC$不是等腰三角形. $\because CD$平分$\angle ACB, \therefore \angle ACD = \angle BCD = \frac{1}{2}\angle ACB = 40^{\circ}$. $\therefore \angle ACD = \angle A = 40^{\circ}$. $\therefore \triangle ACD$为等腰三角形. $\because \angle DCB = \angle A = 40^{\circ}, \angle CBD = \angle ABC, \therefore \triangle BCD \backsim \triangle BAC$. $\therefore CD$是$\triangle ABC$的完美分割线. (2) 由题意, 得$AC = AD = 2$. $\because \triangle BCD \backsim \triangle BAC, \therefore \frac{BC}{BA} = \frac{BD}{BC} = \frac{CD}{AC}$. $\therefore \frac{\sqrt{2}}{BD + 2} = \frac{BD}{\sqrt{2}}$, 解得$BD = \sqrt{3} - 1$(负值舍去). $\therefore \frac{CD}{AC} = \frac{BD}{BC} = \frac{\sqrt{3} - 1}{\sqrt{2}}$. $\therefore CD = \frac{\sqrt{3} - 1}{\sqrt{2}} × 2 = \sqrt{6} - \sqrt{2}$.

答案： 解:
(1) 证明: $\because \angle A = 40^{\circ}, \angle B = 60^{\circ}, \therefore \angle ACB = 80^{\circ}$. $\therefore \triangle ABC$不是等腰三角形. $\because CD$平分$\angle ACB, \therefore \angle ACD = \angle BCD = \frac{1}{2}\angle ACB = 40^{\circ}$. $\therefore \angle ACD = \angle A = 40^{\circ}$. $\therefore \triangle ACD$为等腰三角形. $\because \angle DCB = \angle A = 40^{\circ}, \angle CBD = \angle ABC, \therefore \triangle BCD \backsim \triangle BAC$. $\therefore CD$是$\triangle ABC$的完美分割线.
(2) 由题意, 得$AC = AD = 2$. $\because \triangle BCD \backsim \triangle BAC, \therefore \frac{BC}{BA} = \frac{BD}{BC} = \frac{CD}{AC}$. $\therefore \frac{\sqrt{2}}{BD + 2} = \frac{BD}{\sqrt{2}}$, 解得$BD = \sqrt{3} - 1$(负值舍去). $\therefore \frac{CD}{AC} = \frac{BD}{BC} = \frac{\sqrt{3} - 1}{\sqrt{2}}$. $\therefore CD = \frac{\sqrt{3} - 1}{\sqrt{2}} \times 2 = \sqrt{6} - \sqrt{2}$.