《2025年名校课堂九年级数学全一册人教版广东专版》

5. (2024·广州外国语学校二模)如图，在矩形$ABCD$中，$AB = 3$，$AC = 5$。若$\frac{AF}{FC}=\frac{1}{4}$，则$AE$的长为1。

答案： 5.1

6. (2024·广州海珠区月考)如图，在$\triangle ABC$中，$BD\perp AC$于点$D$，$CE\perp AB$于点$E$，$BD$，$CE$相交于点$O$，连接$DE$。求证：
(1)$\triangle ABD\backsim\triangle ACE$；
(2)$\angle DEC=\angle DBC$。

答案： 6.证明:
(1)$\because BD\perp AC,CE\perp AB,\therefore \angle BDA=\angle CEA=90^{\circ }.\because \angle A=\angle A,\therefore \triangle ABD\backsim \triangle ACE$.
(2)$\because \triangle ABD\backsim \triangle ACE,\therefore \angle OBE=\angle OCD.\because \angle BOE=\angle COD,\therefore \triangle BOE\backsim \triangle COD.\therefore \frac{OB}{OC}=\frac{OE}{OD}.\therefore \frac{OD}{OC}=\frac{OE}{OB}.\because \angle EOD=\angle BOC,\therefore \triangle OED\backsim \triangle OBC.\therefore \angle DEC=\angle DBC.$

7. 如图，在$\triangle ABC$和$\triangle ADE$中，$\angle ACB=\angle AED = 90^{\circ}$，$\angle ABC=\angle ADE$，连接$BD$，$CE$。若$AC:BC = 3:4$，则$BD:CE=$(A)


A.$5:3$
B.$4:3$
C.$\sqrt{5}:2$
D.$2:\sqrt{3}$

答案： 7.A

8. 如图，已知$\angle DAB=\angle EAC$，$\angle ADE=\angle ABC$。求证：
(1)$\triangle ADE\backsim\triangle ABC$；
(2)$\frac{AD}{AE}=\frac{BD}{CE}$。

答案： 8.证明:
(1)$\because \angle DAB=\angle EAC,\therefore \angle DAE=\angle BAC$.又$\because \angle ADE=\angle ABC,\therefore \triangle ADE\backsim \triangle ABC$.
(2)$\because \triangle ADE\backsim \triangle ABC,\therefore \frac{AD}{AE}=\frac{AB}{AC}.\because \angle DAB=\angle EAC,\therefore \triangle ADB\backsim \triangle AEC.\therefore \frac{AD}{AE}=\frac{BD}{CE}.$