《2025年天府前沿九年级数学上册北师大版》

14. (2025·编写)如图,在$\triangle ABC$中,$AB = 4$,$BC = 5$,点$D$,$E分别在BC$,$AC$上,$CD = 2BD$,$CE = 2AE$,$BE交AD于点F$,求$\triangle AFE$面积的最大值。

答案：
[解]如图，连接$DE$

$\because CD=2BD,CE=2AE$,
$\therefore \frac{CD}{BD}=\frac{CE}{AE}=2,\therefore DE// AB$,
$\therefore △CDE\backsim △CBA,\therefore \frac{DE}{BA}=\frac{CD}{CB}=\frac{2}{3}$,
$\therefore \frac{DF}{AF}=\frac{DE}{BA}=\frac{2}{3}$.
$\because DE// AB,\therefore S_{△ABE}=S_{△ABD}$,
$\therefore S_{△AEF}=S_{△BDF},\therefore S_{△AEF}=\frac{2}{5}S_{△ABD}$.
$\because BD=\frac{1}{3}BC=\frac{5}{3},\therefore$ 当$AB⊥BD$时，$△ABD$的面积最大，最大值$=\frac{1}{2}×\frac{5}{3}×4=\frac{10}{3}$,
$\therefore △AEF$的面积的最大值$=\frac{2}{5}×\frac{10}{3}=\frac{4}{3}$.

15. (2025·锦江)如图,在$\triangle ABD$中,$AB = AD$,$\angle BAD = \alpha$。$C是BD$延长线上一动点,连接$AC$,将$AC绕点A按顺时针方向旋转\alpha得到AE$,连接$DE交AC于点F$。
(1)求证：$\angle ACB = \angle E$；
(2)如图1,若$DE // AB$,$DF = 2$,$FE = 7$,求$BD$的长；
(3)如图2,若$F为AC$中点,$\frac{S_{\triangle ADF}}{S_{\triangle ABC}} = $ $\frac{1}{n + 2}$,$CD = 4$,求$AB$的长。(用含$n$的代数式表示)

答案：

(1)[证明]$\because$ 将$AC$绕点$A$按顺时针方向旋转$\alpha$得到$AE$,
$\therefore ∠CAE=\alpha,AC=AE$.
$\because ∠BAD=\alpha$,
$\therefore ∠BAD=∠CAE,\therefore ∠BAC=∠DAE$.
在$△BAC$和$△DAE$中，$\left\{\begin{array}{l} BA=DA,\\ ∠BAC=∠DAE,\\ AC=AE,\end{array}\right.$
$\therefore △BAC\cong △DAE(SAS),\therefore ∠ACB=∠E$.
(2)[解]$\because △BAC\cong △DAE$,
$\therefore ∠ABC=∠ADE,BC=DE=DF+FE=9$.
$\because AB=AD,\therefore ∠B=∠ADB,\therefore ∠ADB=∠ADE$.
$\because DE// AB,\therefore ∠ADE=∠BAD$,
$\therefore ∠B=∠ADB=∠DAB,\therefore AB=AD=BD$.
设$AB=AD=BD=x$,则$CD=9 - x$.
$\because DE// AB,\therefore △CDF\backsim △CBA$,
$\therefore \frac{DF}{AB}=\frac{CD}{CB},\therefore \frac{2}{x}=\frac{9 - x}{9}$,
解得$x=3$或$x=6$.
$\therefore BD$的长为$3$或$6$.
(3)[解]$\because \frac{S_{△ADF}}{S_{△ABC}}=\frac{1}{n+2},△BAC\cong △DAE$,
$\therefore \frac{S_{△ADF}}{S_{△ADE}}=\frac{1}{n+2},\therefore \frac{S_{△ADF}}{S_{△AFE}}=\frac{1}{n+1}$.
$\because F$为$AC$中点，$\therefore S_{△ADF}=S_{△DCF}$,
$\therefore \frac{S_{△DCF}}{S_{△AEF}}=\frac{1}{n+1}$,
由
(1)知，$∠ACB=∠E$.
$\because ∠DFC=∠AFE,\therefore △DFC\backsim △AFE$,
$\therefore \frac{DC}{AE}=\frac{CF}{FE}=\frac{DF}{AF}=\sqrt{\frac{S_{△DFC}}{S_{△AFE}}}=\sqrt{\frac{1}{n+1}}$,
$\therefore \frac{4}{AE}=\sqrt{\frac{1}{n+1}},\therefore AE=4\sqrt{n+1}$,
$\therefore AC=AE=4\sqrt{n+1}$,
$\therefore AF=FC=\frac{1}{2}AC=2\sqrt{n+1}$,
$\therefore \frac{2\sqrt{n+1}}{FE}=\frac{DF}{2\sqrt{n+1}}=\sqrt{\frac{1}{n+1}}$,
$\therefore FE=2n+2,DF=2$.
$\therefore DE=DF+FE=2n+4$.
$\because △BAC\cong △DAE,\therefore BC=DE=2n+4$,
$\because BC=BD+CD,\therefore BD=2n$.
如图，过点$A$作$AM⊥BD$于点$M$，$AN⊥DE$于点$N$.

$\because AB=AD,\therefore BM=DM=\frac{1}{2}BD=n$.
由
(2)知，$∠ADB=∠ADE$.
在$△ADM$和$△ADN$中，$\left\{\begin{array}{l} ∠ADM=∠ADN,\\ ∠AMD=∠AND,\\ AD=AD,\end{array}\right.$
$\therefore △ADM\cong △ADN(AAS),\therefore AM=AN$,
$\therefore DM=DN=n,\therefore EN=DE - DN=n+4$,
$\therefore AN^{2}=AE^{2}-EN^{2}=(4\sqrt{n+1})^{2}-(n+4)^{2}$,
$\therefore AM^{2}=AN^{2}=AE^{2}-EN^{2}=(4\sqrt{n+1})^{2}-(n+4)^{2}=8n - n^{2}$.
在$Rt△ABM$中，根据勾股定理，得$AB=\sqrt{BM^{2}+AM^{2}}=\sqrt{n^{2}+8n - n^{2}}=\sqrt{8n}=2\sqrt{2n}$.