《2025年大联考单元期末测试卷九年级数学全一册沪科版》

14. 教材变式·北师九上 P108T3 如图，在△ABC 中，点 D，E 分别是边 AB，AC 上的点。若 AD = 2，DB = 7，AE = 3，EC = 3，求 DE : BC 的值。

答案： 14.解:$\because AD = 2,DB = 7,AE = 3,EC = 3$,
$\therefore AB = AD + DB = 9,AC = AE + EC = 6$.
$\because \frac {AE} {AB} = \frac {3} {9} = \frac {1} {3},\frac {AD} {AC} = \frac {2} {6} = \frac {1} {3}$,
$\therefore \frac {AE} {AB} = \frac {AD} {AC}$,且$\angle BAC = \angle DAE$,
$\therefore \triangle ABC \sim \triangle AED,\therefore \frac {DE} {BC} = \frac {AD} {AC} = \frac {1} {3}$.(6分)

15. 如图，在△ABC 中，AB = 2，AC = $\sqrt{5}$，点 D 在边 AC 上。若∠ABD = ∠C，求 AD 的长。

答案： 15.解:$\because \angle ABD = \angle C,\angle A = \angle A$,
$\therefore \triangle ABD \sim \triangle ACB,\therefore \frac {AB} {AC} = \frac {AD} {AB}$.(4分)
$\because AB = 2,AC = \sqrt {5},\therefore \frac {2} {\sqrt {5}} = \frac {AD} {2}$,
$\therefore AD = \frac {4 \sqrt {5}} {5}$.(8分)

16. 如图，四边形 ABCD 是正方形，点 G 为边 CD 上一点，连接 AG 并延长，交 BC 的延长线于点 F，连接 BD 交 AF 于点 E，连接 EC。求证：△EGC∽△ECF。

答案： 16.证明:$\because$四边形ABCD是正方形，
$\therefore AD = CD,\angle ADE = \angle CDE = 45 ^ { \circ }$.
$\because DE = DE$,
$\therefore \triangle ADE \cong \triangle CDE(SAS)$,
$\therefore \angle DAE = \angle DCE$.(4分)
$\because$四边形ABCD是正方形,点F在BC的延长线上，
$\therefore AD // BF$,
$\therefore \angle DAE = \angle F$,
$\therefore \angle F = \angle DCE$.
又$\because \angle CEG = \angle FEC$,
$\therefore \triangle EGC \sim \triangle ECF$.(8分)