《2025年零障碍导教导学案九年级数学全一册北师大版》

5. (2024·龙岗区期中) 如图，已知 $\angle 1 = \angle 2$，那么添加下列的一个条件后，仍无法判定 $\triangle ABC \backsim \triangle ADE$ 的是 (C)


A. $\frac{AB}{AD} = \frac{AC}{AE}$
B. $\angle B = \angle D$
C. $\frac{AB}{AD} = \frac{BC}{DE}$
D. $\angle C = \angle AED$

答案： C

6. 如图，$AD$，$BC$ 相交于点 $O$，$P$ 为 $AB$，$CD$ 延长线的交点，且 $PA \cdot PB = PC \cdot PD$.
求证：$\triangle PAD \backsim \triangle PCB$.

答案： 证明:$\because PA \cdot PB = PC \cdot PD$,
$\therefore \frac{PA}{PC} = \frac{PD}{PB}$.
又$\because \angle P = \angle P$,
$\therefore \triangle PAD \backsim \triangle PCB$.

7. 【原创题】如图，$\angle A = \angle D = \angle CBE = \alpha$，点 $A$，$B$，$D$ 在同一直线上.
(1) 若 $\alpha = 60^{\circ}$，求证：图中两个三角形相似；
(2) 若 $AC = DE = 2$，$AD = 5$，求 $AB$ 的长.
(1) 证明:$\because \angle ABC + \angle C = 180^{\circ} -$$\angle A = 120^{\circ}$,$\angle ABC + \angle DBE = 180^{\circ} - \angle CBE$$= 120^{\circ}$,$\therefore \angle C = \angle DBE$.又$\because \angle A = \angle D$,$\therefore \triangle ABC \backsim \triangle DEB$.
(2)解:在$\triangle ABC$中.$\because \angle A = \alpha$,$\therefore \angle C + \angle ABC = 180^{\circ} - \alpha$.$\because \angle CBE = \alpha$,$\therefore \angle EBD + \angle ABC = 180^{\circ} - \alpha$.$\therefore \angle C = \angle EBD$.又$\because \angle A = \angle D = \alpha$,$\therefore \triangle ABC \backsim \triangle DEB$.$\therefore \frac{AC}{DB} = \frac{AB}{DE}$.设$AB = x$,则$BD = 5 - x$.$\therefore \frac{2}{5 - x} = \frac{x}{2}$.$\therefore 5x - x^{2} = 4$.$\therefore x^{2} - 5x + 4 = 0$,解得$x_{1} = 1$,$x_{2} = 4$.$\therefore AB = $1或4.

答案：
(1) 证明:$\because \angle ABC + \angle C = 180^{\circ} -$
$\angle A = 120^{\circ}$,
$\angle ABC + \angle DBE = 180^{\circ} - \angle CBE$
$= 120^{\circ}$,
$\therefore \angle C = \angle DBE$.
又$\because \angle A = \angle D$,
$\therefore \triangle ABC \backsim \triangle DEB$.
(2)解:在$\triangle ABC$中.
$\because \angle A = \alpha$,
$\therefore \angle C + \angle ABC = 180^{\circ} - \alpha$.
$\because \angle CBE = \alpha$,
$\therefore \angle EBD + \angle ABC = 180^{\circ} - \alpha$.
$\therefore \angle C = \angle EBD$.
又$\because \angle A = \angle D = \alpha$,
$\therefore \triangle ABC \backsim \triangle DEB$.
$\therefore \frac{AC}{DB} = \frac{AB}{DE}$.
设$AB = x$,则$BD = 5 - x$.
$\therefore \frac{2}{5 - x} = \frac{x}{2}$.
$\therefore 5x - x^{2} = 4$.
$\therefore x^{2} - 5x + 4 = 0$,
解得$x_{1} = 1$,$x_{2} = 4$.
$\therefore AB = 1$或$AB = 4$.

8. 如图，在 $\triangle ABC$ 中，$\angle ACB = 90^{\circ}$，$CA = CB = 2\sqrt{2}$. $D$，$E$ 为 $AB$ 上两点，且 $\angle DCE = 45^{\circ}$.
(1) 求证：$\triangle ACE \backsim \triangle BDC$；
(2) 若 $AD = 1$，求 $DE$ 的长.


(1)证明:$\because \angle ACB = 90^{\circ}$,
$CA = CB$,
$\therefore \angle A = \angle B$
$= \frac{1}{2}(180^{\circ} - 90^{\circ})$
$= 45^{\circ}$.
又$\because \angle CDB = \angle A + \angle ACD$
$= 45^{\circ} + \angle ACD$
$= \angle ACE$,
$\therefore \triangle ACE \backsim \triangle BDC$.
(2)解:由勾股定理,得
$AB = \sqrt{(2\sqrt{2})^{2} + (2\sqrt{2})^{2}} = 4$.
设$DE$的长为$x$.
$\because AD = 1$,
$\therefore BD = 3$,$AE = 1 + x$.
$\because \triangle ACE \backsim \triangle BDC$,
$\therefore \frac{AC}{BD} = \frac{AE}{BC}$,即$\frac{2\sqrt{2}}{3} = \frac{1 + x}{2\sqrt{2}}$,
解得$x = $$\frac{5}{3}$,即$DE = $$\frac{5}{3}$.

答案：
(1)证明:$\because \angle ACB = 90^{\circ}$,
$CA = CB$,
$\therefore \angle A = \angle B$
$= \frac{1}{2}(180^{\circ} - 90^{\circ})$
$= 45^{\circ}$.
又$\because \angle CDB = \angle A + \angle ACD$
$= 45^{\circ} + \angle ACD$
$= \angle ACE$,
$\therefore \triangle ACE \backsim \triangle BDC$.
(2)解:由勾股定理,得
$AB = \sqrt{(2\sqrt{2})^{2} + (2\sqrt{2})^{2}} = 4$.
设$DE$的长为$x$.
$\because AD = 1$,
$\therefore BD = 3$,$AE = 1 + x$.
$\because \triangle ACE \backsim \triangle BDC$,
$\therefore \frac{AC}{BD} = \frac{AE}{BC}$,即$\frac{2\sqrt{2}}{3} = \frac{1 + x}{2\sqrt{2}}$,
解得$x = \frac{5}{3}$,即$DE = \frac{5}{3}$.