《2025年名校课堂九年级数学全一册湘教版广西专版》

7. 如图,在$\triangle ABC$中,点 D,E 分别在边 AC,AB 上,BD,CE 相交于点 O,且$\frac {EO}{BO}=\frac {DO}{CO}$. 求证:$\triangle ADE\backsim \triangle ABC$.

答案： 证明：$\because \frac{EO}{BO}=\frac{DO}{CO}$，$\angle DOE=\angle COB$，$\therefore \triangle DOE\backsim \triangle COB$，$\frac{EO}{DO}=\frac{BO}{CO}$．$\therefore \angle DEO=\angle CBO$．$\because \angle BOE=\angle COD$，$\therefore \triangle BOE\backsim \triangle COD$．$\therefore \angle EBO=\angle DCO$．$\because \angle ADE=\angle DCO+\angle DEO$，$\angle ABC=\angle EBO+\angle CBO$，$\therefore \angle ADE=\angle ABC$．又$\because \angle A=\angle A$，$\therefore \triangle ADE\backsim \triangle ABC$．

8. 如图,在$\triangle ABC$中,$∠ACB=90^{\circ }$,CD 是斜边 AB 上的高,$AD=9,BD=4$,那么$CD=$6,$AC=$$3\sqrt{13}$.

答案： 6 $3\sqrt{13}$

9. 如图,点$P_{1},P_{2},P_{3},P_{4}$均在坐标轴上,且$P_{1}P_{2}⊥P_{2}P_{3},P_{2}P_{3}⊥P_{3}P_{4}$. 若点$P_{1},P_{2}$的坐标分别为$(0,-1),(-2,0)$,则点$P_{4}$的坐标为$(8,0)$.

答案： $(8,0)$

10. 如图,在$\triangle ABC$和$\triangle ADE$中,$∠ACB=∠AED=90^{\circ },∠ABC=∠ADE$,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： A

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

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