《2025年零障碍导教导学案九年级数学全一册北师大版》

1. 例 如图，正方形 $ABCD$ 的边长为 $4$，$BF = 1$，$E$ 为 $AB$ 的中点.
(1) 证明图中一对相似三角形：$\triangle ADE \backsim \triangle BEF$.
证明如下:
$\because E$为$AB$的中点,
$\therefore AE = BE = 2$.
$\therefore \frac{AE}{BF} = \frac{2}{1}$,$\frac{AD}{BE} = \frac{4}{2} = \frac{2}{1}$.
$\therefore \frac{AE}{BF} = \frac{AD}{BE} = 2$.
又$\because \angle A = \angle B$,
$\therefore \triangle ADE \backsim \triangle BEF$.
(2) 求证：$DE \perp EF$.
证明:$\because \triangle ADE \backsim \triangle BEF$,
$\therefore \angle ADE = \angle BEF$.
$\therefore \angle DEF = 180^{\circ} - \angle AED - \angle BEF$
$= 180^{\circ} - \angle AED - \angle ADE$
$= \angle A = 90^{\circ}$.
$\therefore DE \perp EF$.

答案：
(1)解:$\triangle ADE \backsim \triangle BEF$.
证明如下:
$\because E$为$AB$的中点,
$\therefore AE = BE = 2$.
$\therefore \frac{AE}{BF} = \frac{2}{1}$,$\frac{AD}{BE} = \frac{4}{2} = \frac{2}{1}$.
$\therefore \frac{AE}{BF} = \frac{AD}{BE} = 2$.
又$\because \angle A = \angle B$,
$\therefore \triangle ADE \backsim \triangle BEF$.
(2)证明:$\because \triangle ADE \backsim \triangle BEF$,
$\therefore \angle ADE = \angle BEF$.
$\therefore \angle DEF = 180^{\circ} - \angle AED - \angle BEF$
$= 180^{\circ} - \angle AED - \angle ADE$
$= \angle A = 90^{\circ}$.
$\therefore DE \perp EF$.

2. 如图，在 $\triangle ABC$ 中，$D$ 为 $AB$ 上一点，$AD = 4$，$BD = 5$，$AC = 6$. 求证：
(1) $\triangle ABC \backsim \triangle ACD$；
(2) $BC \cdot AD = AC \cdot CD$.

答案： 证明:
(1)$\because AD = 4$,$BD = 5$,
$AC = 6$,
$\therefore AB = AD + BD = 4 + 5 = 9$.
$\therefore \frac{AD}{AC} = \frac{4}{6} = \frac{2}{3}$,$\frac{AC}{AB} = \frac{6}{9} = \frac{2}{3}$.
$\therefore \frac{AD}{AC} = \frac{AC}{AB} = \frac{2}{3}$.
又$\because \angle A = \angle A$,
$\therefore \triangle ABC \backsim \triangle ACD$.
(2)由
(1)知$\triangle ABC \backsim \triangle ACD$,
$\therefore \frac{BC}{CD} = \frac{AC}{AD}$,
即$BC \cdot AD = AC \cdot CD$.

3. 例 (2024·龙华区期中) 如图，在矩形 $ABCD$ 中，$E$ 为 $BC$ 上一点，$DF \perp AE$ 于点 $F$.
(1) 求证：$\triangle ABE \backsim \triangle DFA$；
(2) 若 $AB = 3$，$AD = 6$，$BE = 4$，求 $DF$ 的长.


(1)证明:$\because$四边形$ABCD$是矩形,$\therefore AD // BC$,$\angle B = 90^{\circ}$.$\therefore \angle AEB = \angle DAF$.$\because DF \perp AE$,$\therefore \angle DFA = 90^{\circ}$.$\therefore \angle B = \angle DFA$.$\therefore \triangle ABE \backsim \triangle DFA$.
(2)解:$\because AB = 3$,$BE = 4$,$\therefore$由勾股定理,得$AE =$5.$\because \triangle ABE \backsim \triangle DFA$,$\therefore \frac{AE}{DA} = \frac{AB}{DF}$,即$\frac{5}{6} = \frac{3}{DF}$.$\therefore DF =$3.6.

答案：
(1)证明:$\because$四边形$ABCD$
是矩形,
$\therefore AD // BC$,$\angle B = 90^{\circ}$.
$\therefore \angle AEB = \angle DAF$.
$\because DF \perp AE$,
$\therefore \angle DFA = 90^{\circ}$.
$\therefore \angle B = \angle DFA$.
$\therefore \triangle ABE \backsim \triangle DFA$.
(2)解:$\because AB = 3$,$BE = 4$,
$\therefore$由勾股定理,得$AE = 5$.
$\because \triangle ABE \backsim \triangle DFA$,
$\therefore \frac{AE}{DA} = \frac{AB}{DF}$,即$\frac{5}{6} = \frac{3}{DF}$.
$\therefore DF = 3.6$.

4. 如图，在平行四边形 $ABCD$ 中，过点 $A$ 作 $AE \perp BC$，垂足为 $E$，连接 $DE$，点 $F$ 为 $DE$ 上一点，且 $\angle AFE = \angle B$.
(1) 求证：$\triangle ADF \backsim \triangle DEC$；
(2) 若 $AE = 6$，$AD = 8$，$AB = 7$，求 $AF$ 的长.

(1)证明:在平行四边形$ABCD$中,
$AB // DC$,
$\therefore \angle C + \angle B = 180^{\circ}$.
又$\because \angle AFD + \angle AFE = 180^{\circ}$,
$\angle AFE = \angle B$,
$\therefore \angle AFD = \angle C$.
$\because AD // BC$,
$\therefore \angle ADF = \angle DEC$.
$\therefore \triangle ADF \backsim \triangle DEC$.
(2)解:$\because$四边形$ABCD$是平行四边形,
$\therefore AD // BC$,$AB = CD = 7$.
$\because AE \perp BC$,
$\therefore AE \perp AD$.
$\therefore DE = \sqrt{AE^{2} + AD^{2}}$
$= \sqrt{6^{2} + 8^{2}} = 10$.
由(1)可知$\triangle ADF \backsim \triangle DEC$,
$\therefore \frac{AF}{DC} = \frac{AD}{DE}$,即$\frac{AF}{7} = \frac{8}{10}$.
$\therefore AF =$$\frac{28}{5}$.

答案：
(1)证明:在平行四边形$ABCD$中,
$AB // DC$,
$\therefore \angle C + \angle B = 180^{\circ}$.
又$\because \angle AFD + \angle AFE = 180^{\circ}$,
$\angle AFE = \angle B$,
$\therefore \angle AFD = \angle C$.
$\because AD // BC$,
$\therefore \angle ADF = \angle DEC$.
$\therefore \triangle ADF \backsim \triangle DEC$.
(2)解:$\because$四边形$ABCD$是平行四边形,
$\therefore AD // BC$,$AB = CD = 7$.
$\because AE \perp BC$,
$\therefore AE \perp AD$.
$\therefore DE = \sqrt{AE^{2} + AD^{2}}$
$= \sqrt{6^{2} + 8^{2}} = 10$.
由
(1)可知$\triangle ADF \backsim \triangle DEC$,
$\therefore \frac{AF}{DC} = \frac{AD}{DE}$,即$\frac{AF}{7} = \frac{8}{10}$.
$\therefore AF = \frac{28}{5}$.