《2025年名师帮同步学案九年级数学全一册人教版》

一、新课学习
相似三角形的判定2：三边成比例的两个三角形相似(如图所示).
∵$ \frac{AB}{A'B'} = \frac{AC}{A'C'} = \frac{BC}{B'C'} $,
∴ $ \triangle ABC \backsim \triangle A'B'C' $.

相似三角形的判定3：两边成比例且夹角相等的两个三角形相似(如图所示).
∵$ \frac{AB}{A'B'} = \frac{AC}{A'C'} $,
$ \angle A = \angle A' $,
∴ $ \triangle ABC \backsim \triangle A'B'C' $.

答案： 一、新课学习
$ \frac{AB}{A'B'} = \frac{AC}{A'C'} = \frac{BC}{B'C'} $
$ \frac{AB}{A'B'} = \frac{AC}{A'C'} \quad \angle A = \angle A' $

【例题1】如图,根据所给条件,判断$ \triangle ABC 和 \triangle DEF $是否相似,并说明理由.

解：相似.理由如下：
$\because \frac{AB}{DE} = \frac{14}{7} = 2, \frac{BC}{EF} = \frac{12}{6} = 2,$
$\frac{AC}{DF} = \frac{10}{5} = 2,$
$\therefore \frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} = 2,$
$\therefore \triangle ABC \sim \triangle DEF.$

答案： 【例题1】解：相似.理由如下：
$\because \frac{AB}{DE} = \frac{14}{7} = 2, \frac{BC}{EF} = \frac{12}{6} = 2,$
$\frac{AC}{DF} = \frac{10}{5} = 2,$
$\therefore \frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} = 2,$
$\therefore \triangle ABC \sim \triangle DEF.$

【变式1】如图,在正方形网格中,每个小正方形的边长为1,求两个三角形的各边边长,并证明：$ \triangle ABC \backsim \triangle A_1B_1C_1 $.

解：$ \because $每个小正方形的边长为1，
$\therefore AB = $$\sqrt{5}$,$ AC = $$\sqrt{10}$,$ BC = $5,$ A_1B_1 = $$\sqrt{2}$,
$A_1C_1 = $2,$ B_1C_1 = $$\sqrt{10}$;
证明：$\because \frac{A_1B_1}{AB} = \frac{\sqrt{2}}{\sqrt{5}} = $$\frac{\sqrt{10}}{5}$,$ \frac{A_1C_1}{AC} = \frac{2}{\sqrt{10}} = $$\frac{\sqrt{10}}{5}$,$ \frac{B_1C_1}{BC} = $$\frac{\sqrt{10}}{5}$,
$\therefore \frac{A_1B_1}{AB} = \frac{A_1C_1}{AC} = \frac{B_1C_1}{BC},$
$\therefore \triangle ABC \sim \triangle A_1B_1C_1.$

答案： 【变式1】解：$ \because $每个小正方形的边长为1，
$\therefore AB = \sqrt{5}, AC = \sqrt{10}, BC = 5, A_1B_1 = \sqrt{2},$
$A_1C_1 = 2, B_1C_1 = \sqrt{10};$
证明：$\because \frac{A_1B_1}{AB} = \frac{\sqrt{2}}{\sqrt{5}} = \frac{\sqrt{10}}{5}, \frac{A_1C_1}{AC} = \frac{2}{\sqrt{10}} =$
$\frac{\sqrt{10}}{5}, \frac{B_1C_1}{BC} = \frac{\sqrt{10}}{5},$
$\therefore \frac{A_1B_1}{AB} = \frac{A_1C_1}{AC} = \frac{B_1C_1}{BC},$
$\therefore \triangle ABC \sim \triangle A_1B_1C_1.$

【例题2】如图,点A,B,C,D均在边长为1的小正方形网格的格点上,连接AD. 求证：$ \triangle ABD \backsim \triangle CBA $.

解：根据勾股定理，得$ AB = \sqrt{2^2 + 1^2} = $$\sqrt{5}$,$ BD = $1,$ BC = $5,$\therefore \frac{AB}{BC} = $$\frac{\sqrt{5}}{5}$,$ \frac{BD}{AB} = \frac{1}{\sqrt{5}} = $$\frac{\sqrt{5}}{5}$,$\therefore \frac{AB}{BC} = \frac{BD}{AB},$又$ \because \angle ABD = $$\angle CBA$,$\therefore \triangle ABD \sim \triangle CBA.$

答案： 【例题2】解：根据勾股定理，得$ AB = \sqrt{2^2 + 1^2} = $
$\sqrt{5}, BD = 1, BC = 5,$
$\therefore \frac{AB}{BC} = \frac{\sqrt{5}}{5}, \frac{BD}{AB} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5},$
$\therefore \frac{AB}{BC} = \frac{BD}{AB},$
又$ \because \angle ABD = \angle CBA, $
$\therefore \triangle ABD \sim \triangle CBA.$

【变式2】如图,在$ \triangle ABC $中,CD是AB边上的高,且$ CD^2 = AD \cdot BD $.
(1) 求证：$ \triangle ACD \backsim \triangle CBD $;
证明：$ \because CD $是$ AB $边上的高，
$\therefore \angle ADC = \angle CDB = 90^\circ,$
$\because CD^2 = AD \cdot BD, \therefore \frac{AD}{CD} = \frac{CD}{BD},$
$\therefore \triangle ACD \sim \triangle CBD.$
(2) 若$ AC = 3 $,$ BC = 4 $,求AB的长.
解：由(1)，得$ \triangle ACD \sim \triangle CBD. $
$\therefore \angle ACD = \angle B.$
$\because \angle BCD + \angle B = 180^\circ - \angle CDB = 90^\circ,$
$\therefore \angle BCD + \angle ACD = 90^\circ, \text{即} \angle ACB = 90^\circ,$
在$ \text{Rt} \triangle ABC $中，$ AB = \sqrt{AC^2 + BC^2} = \sqrt{3^2 + 4^2} = $5.

答案： 【变式2】
(1)证明：$ \because CD $是$ AB $边上的高，
$\therefore \angle ADC = \angle CDB = 90^\circ,$
$\because CD^2 = AD \cdot BD, \therefore \frac{AD}{CD} = \frac{CD}{BD},$
$\therefore \triangle ACD \sim \triangle CBD.$
(2)解：由
(1)，得$ \triangle ACD \sim \triangle CBD. $
$\therefore \angle ACD = \angle B.$
$\because \angle BCD + \angle B = 180^\circ - \angle CDB = 90^\circ,$
$\therefore \angle BCD + \angle ACD = 90^\circ, \text{即} \angle ACB = 90^\circ,$
在$ \text{Rt} \triangle ABC $中，$ AB = \sqrt{AC^2 + BC^2} = \sqrt{3^2 + 4^2} = 5. $

1. 如图,以点A,B,C为顶点的三角形与以点D,E,F为顶点的三角形相似,则$ \triangle ABC 与 \triangle DEF $的相似比为(B)

A.$ 1 : 2 $
B.$ 2 : 1 $
C.$ 3 : 4 $
D.$ 4 : 3 $

答案： 1. B

2. (人教九下P42教材改编)如图,在大小为$ 4 × 4 $的正方形网格中,是相似三角形的是(C)

A.①②
B.②③
C.①③
D.②④

答案： 2. C

3. 如图,点D在$ \triangle ABC $的AB边上,$ AD = 2 $,$ BD = 4 $,$ AC = 2\sqrt{3} $.

(1) 求证：$ \triangle ACD \backsim \triangle ABC $;
证明：$\because \frac{AD}{AC} = \frac{2}{2\sqrt{3}} = \frac{\sqrt{3}}{3}, \frac{AC}{AB} = \frac{2\sqrt{3}}{2 + 4} = \frac{\sqrt{3}}{3},$$\therefore \frac{AD}{AC} = \frac{AC}{AB}.$又$ \because \angle A = \angle A, $$\therefore \triangle ACD \sim \triangle ABC;$
(2) 求$ CD : BC $的值.$\frac{\sqrt{3}}{3}$

答案： 3.
(1)证明：$\because \frac{AD}{AC} = \frac{2}{2\sqrt{3}} = \frac{\sqrt{3}}{3}, \frac{AC}{AB} = \frac{2\sqrt{3}}{2 + 4} = \frac{\sqrt{3}}{3},$
$\therefore \frac{AD}{AC} = \frac{AC}{AB}.$
又$ \because \angle A = \angle A, $
$\therefore \triangle ACD \sim \triangle ABC;$
(2)解：$\because \triangle ACD \sim \triangle ABC,$
$\therefore \frac{CD}{BC} = \frac{AC}{AB} = \frac{\sqrt{3}}{3}.$

4. 如图,在$ \triangle ABC $中,点D,E,F分别是AB,BC,CA的中点. 求证：$ \triangle ABC \backsim \triangle EFD $.

证明：$ \because D, E, F $分别是$ AB, BC, CA $的中点，
$\therefore \frac{DF}{BC} = \frac{DE}{AC} = \frac{EF}{AB} = \frac{1}{2},$
$\therefore$$\triangle ABC \sim \triangle EFD$.

答案： 4. 证明：$ \because D, E, F $分别是$ AB, BC, CA $的中点，
$\therefore \frac{DF}{BC} = \frac{DE}{AC} = \frac{EF}{AB} = \frac{1}{2},$
$\therefore \triangle ABC \sim \triangle EFD.$

5. 如图,在正方形ABCD中,点E是边AD的中点,点F在边CD上,且$ CD = 4DF $,连接BE,EF.
求证：(1) $ \triangle ABE \backsim \triangle DEF $;
(2) $ \angle BEF = 90^\circ $.

答案： 5. 证明：
(1)设$ AB = 4k. $
$\because \text{四边形} ABCD \text{是正方形},$
$\therefore AB = AD = CD = 4k, \angle A = \angle D = 90^\circ.$
$\because CD = 4DF, \text{点} E \text{是边} AD \text{的中点},$
$\therefore DF = k, AE = DE = 2k,$
$\therefore \frac{AE}{DF} = \frac{AB}{DE} = 2,$
$\therefore \triangle ABE \sim \triangle DEF.$
(2)$\because \triangle ABE \sim \triangle DEF,$
$\therefore \angle ABE = \angle DEF.$
$\because \angle ABE + \angle AEB = 90^\circ,$
$\therefore \angle DEF + \angle AEB = 90^\circ.$
$\therefore \angle BEF = 180^\circ - (\angle DEF + \angle AEB) = 90^\circ.$

6. 如图,在$ \triangle ABC $中,$ \angle C = 90^\circ $,$ BC = 8 \text{ cm} $,$ AC = 6 \text{ cm} $,点Q从点B出发,以$ 2 \text{ cm/s} $的速度沿BC方向移动,点P从点C出发,以$ 1 \text{ cm/s} $的速度沿CA方向移动,若点Q,P分别同时从点B,C出发,经过$\frac{12}{5}$或$\frac{32}{11}$秒后,以点C,P,Q为顶点的三角形与$ \triangle CBA $相似.

答案： 6. 解：设经过$ x \text{ s} $后，两个三角形相似，则$ CQ = $
$(8 - 2x) \text{ cm}, CP = x \text{ cm}.$
$\because \angle C = \angle C = 90^\circ,$
$\therefore \text{当} \frac{CQ}{CB} = \frac{CP}{CA} \text{或} \frac{CQ}{CA} = \frac{CP}{CB} \text{时，两个三角形}$
相似.
当$ \frac{CQ}{CB} = \frac{CP}{CA} $时，$ \frac{8 - 2x}{8} = \frac{x}{6} $，解得$ x = \frac{12}{5} $；
当$ \frac{CQ}{CA} = \frac{CP}{CB} $时，$ \frac{8 - 2x}{6} = \frac{x}{8} $，解得$ x = \frac{32}{11} $.
$\therefore \text{经过} \frac{12}{5} \text{ s} \text{或} \frac{32}{11} \text{ s} \text{后，两个三角形相似}.$