《2025年精英新课堂九年级数学全一册北师大版贵州专版》

5. 如图，在$\triangle ABC$中，点$D$在边$BC$上，$E$为$AD$上一点，连接$CE$. 若$\angle DAC = \angle B$，$CD = CE$，求证：$CD\cdot BD = AD\cdot AE$.

答案： 证明：$\because CD = CE$，$\therefore \angle CED=\angle CDE$.$\therefore \angle CEA=\angle ADB$.$\because \angle DAC=\angle B$，$\therefore \triangle CEA\backsim\triangle ADB$.$\therefore \frac{CE}{AD}=\frac{AE}{BD}$.$\therefore \frac{CD}{AD}=\frac{AE}{BD}$.$\therefore CD\cdot BD = AD\cdot AE$.

6. 如图，在$\triangle ABC$中，$AB = AC = 16$，$D$是$BC$上一点，$E$是$AC$上一点，$\angle ADE = \angle C$. 若$CE:BD = 3:4$，则$CD$的长为(　　　)
A. 12
B. 10
C. 8
D. 6

答案： A

7. 如图，在正方形$ABCD$中，$AB = 9$，$E$为$BC$上一点，过点$E$作$EF\perp AE$，交$CD$于点$F$，连接$AF$.
(1)求证:$\triangle ABE\sim\triangle ECF$;
(2)当$BE = 3$时，求$CF$的长.

答案：
(1) 证明：$\because$四边形$ABCD$是正方形，$\therefore \angle B=\angle C = 90^{\circ}$.$\therefore \angle BAE+\angle AEB = 90^{\circ}$.$\because EF\perp AE$，$\therefore \angle AEF = 90^{\circ}$.$\therefore \angle AEB+\angle CEF = 90^{\circ}$.$\therefore \angle BAE=\angle CEF$.$\therefore \triangle ABE\backsim\triangle ECF$.
(2) 解：$\because$四边形$ABCD$是正方形，$\therefore BC = AB = 9$.$\because BE = 3$，$\therefore EC = BC - BE = 6$.$\because \triangle ABE\backsim\triangle ECF$，$\therefore \frac{AB}{EC}=\frac{BE}{CF}$.$\therefore \frac{9}{6}=\frac{3}{CF}$.$\therefore CF = 2$.

8. 如图，$\triangle ABC$和$\triangle CEF$均为等腰直角三角形，$\angle ABC = \angle CFE = 90^{\circ}$，点$E$在$\triangle ABC$内，$\angle CAE+\angle CBE = 90^{\circ}$，连接$BF$.
(1)求证:$\triangle CAE\sim\triangle CBF$;
(2)若$BE = 1$，$AE = 2$，求$CE$的长.

答案：
(1) 证明：$\because \triangle ABC$和$\triangle CEF$均为等腰直角三角形，$\therefore \angle ACB=\angle ECF = 45^{\circ}$.$\therefore \angle ACB-\angle ECB=\angle ECF-\angle ECB$，即$\angle ACE=\angle BCF$. 易得$\frac{AC}{BC}=\frac{CE}{CF}=\sqrt{2}$，$\therefore \triangle CAE\backsim\triangle CBF$.
(2) 解：$\because \triangle CAE\backsim\triangle CBF$，$\therefore \angle CAE=\angle CBF$，$\frac{AE}{BF}=\frac{AC}{BC}=\sqrt{2}$.$\because AE = 2$，$\therefore BF=\sqrt{2}$.$\because \angle CAE+\angle CBE = 90^{\circ}$，$\therefore \angle CBF+\angle CBE = 90^{\circ}$，即$\angle EBF = 90^{\circ}$. 在$Rt\triangle EBF$中，$EF=\sqrt{BE^{2}+BF^{2}}=\sqrt{3}$.$\because \triangle CEF$是等腰直角三角形，$\therefore CF = EF=\sqrt{3}$.$\therefore CE=\sqrt{EF^{2}+CF^{2}}=\sqrt{6}$.