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1. 教材P96例4变式 [2025梧州期末]若$△ ABC$的三边长分别是3,5,6,则与$△ ABC$相似的$△ DEF$的边可能是（
A.$DE=6,DF=8,EF=10$
B.$DE=9,EF=18,DF=25$
C.$DE=1,EF=2,DF=2.5$
D.$DE=6,DF=10,EF=12$
D
）A.$DE=6,DF=8,EF=10$
B.$DE=9,EF=18,DF=25$
C.$DE=1,EF=2,DF=2.5$
D.$DE=6,DF=10,EF=12$
答案
1 D $△ ABC$ 的三边长分别是 3,5,6,当 $DE = 6, DF = 8$,$EF = 10$ 时,$\frac{3}{6} ≠ \frac{5}{8} ≠ \frac{6}{10}$,故选项 A 不符合题意;当 $DE = 9$,$EF = 18, DF = 25$ 时,$\frac{3}{9} ≠ \frac{5}{18} ≠ \frac{6}{25}$,故选项 B 不符合题意;当$DE = 1,EF = 2,DF = 2.5$ 时,$\frac{3}{1} ≠ \frac{5}{2} ≠ \frac{6}{2.5}$,故选项 C 不符合题意;当 $DE = 6, DF = 10, EF = 12$ 时, $\frac{3}{6}=\frac{5}{10}=\frac{6}{12}$,故选项 D符合题意.
2. [2026蚌埠一模]如图,每个小正方形的边长均为1,则下列图形中的三角形(阴影部分)与$△ A_1B_1C_1$相似的是 (
A
B
C D
B
)A
B
C D
答案
2 B 由题图,得 $A_1B_1 = 1$,$B_1C_1 = \sqrt{2}$,$A_1C_1 = \sqrt{5}$,$\therefore A_1B_1:B_1C_1:A_1C_1 = 1:\sqrt{2}:\sqrt{5}$. A 项,三边(由短到长)之比为 $\sqrt{2}:\sqrt{5}:3$,不符合题意;B 项,三边(由短到长)之比为 $\sqrt{2}:2:\sqrt{10} = 1:\sqrt{2}:\sqrt{5}$,符合题意;C 项,三边(由短到长)之比为 $1:\sqrt{5}:2\sqrt{2}$,不符合题意;D 项,三边(由短到长)之比为 $2:\sqrt{5}:\sqrt{13}$,不符合题意.
变式[2025濮阳期末]如图,在由小正方形组成的网格中有6个三角形:①$△ ABC$, ②$△ CDB$, ③$△ DEB$, ④$△ FBG$, ⑤$△ HGF$, ⑥$△ EKF$. 则在②~⑥中,与①相似的三角形有
③④⑤
. (填序号)答案
变式 ③④⑤ ①中三边(由短到长)之比为 $1:\sqrt{2}:\sqrt{5}$,② ~ ⑥中三边(由短到长)之比与①中相同的有③④⑤,故与①相似的三角形是③④⑤.
3. 教材 P97T3 变式·一题多解 如图,已知 O 是△ABC内一点,D,E,F分别是OA,OB,OC的中点.求证:△DEF∽△ABC.
答案
3 证明:
通解 $\because D,E$分别为 $OA,OB$ 的中点,
$\therefore DE$ 是 $△ OAB$ 的中位线,$\therefore \frac{DE}{AB}=\frac{1}{2}$.
同理可得 $\frac{EF}{BC}=\frac{1}{2}$,$\frac{DF}{AC}=\frac{1}{2}$,
$\therefore \frac{DE}{AB}=\frac{EF}{BC}=\frac{DF}{AC}$,$\therefore △ DEF ∽ △ ABC$.
另解 $\because D,E$ 分别为 $OA,OB$ 的中点,
$\therefore DE // AB$,$\therefore ∠ OED = ∠ ABO$,$∠ ODE = ∠ OAB$.
同理可得 $∠ ODF = ∠ OAC$,$∠ OEF = ∠ OBC$,
$\therefore ∠ DEF = ∠ ABC$,$∠ EDF = ∠ BAC$,
$\therefore △ DEF ∽ △ ABC$.
通解 $\because D,E$分别为 $OA,OB$ 的中点,
$\therefore DE$ 是 $△ OAB$ 的中位线,$\therefore \frac{DE}{AB}=\frac{1}{2}$.
同理可得 $\frac{EF}{BC}=\frac{1}{2}$,$\frac{DF}{AC}=\frac{1}{2}$,
$\therefore \frac{DE}{AB}=\frac{EF}{BC}=\frac{DF}{AC}$,$\therefore △ DEF ∽ △ ABC$.
另解 $\because D,E$ 分别为 $OA,OB$ 的中点,
$\therefore DE // AB$,$\therefore ∠ OED = ∠ ABO$,$∠ ODE = ∠ OAB$.
同理可得 $∠ ODF = ∠ OAC$,$∠ OEF = ∠ OBC$,
$\therefore ∠ DEF = ∠ ABC$,$∠ EDF = ∠ BAC$,
$\therefore △ DEF ∽ △ ABC$.
4. 如图，$△ ABC$各顶点的坐标分别为$A(3,0)$，$B(0,4)$，$C(4,2)$，过点$C$作$CD ⊥ x$轴，垂足为$D$.求证:$△ ABC ∽ △ ACD$.
答案
4 证明:$\because A(3,0)$,$B(0,4)$,$C(4,2)$,$CD ⊥ x$轴,
$\therefore OA = 3$,$OB = 4$,$OD = 4$,$CD = 2$,
$\therefore AD = 4 - 1 = 1$,
$\therefore AB = \sqrt{3^2 + 4^2} = 5$,$BC = \sqrt{4^2 + (4 - 2)^2} = 2\sqrt{5}$,
$AC = \sqrt{(3 - 4)^2 + 2^2} = \sqrt{5}$.
$\because \frac{AB}{AC} = \frac{5}{\sqrt{5}} = \sqrt{5}$,$\frac{BC}{CD} = \frac{2\sqrt{5}}{2} = \sqrt{5}$,$\frac{AC}{AD} = \frac{\sqrt{5}}{1} = \sqrt{5}$,
$\therefore \frac{AB}{AC} = \frac{BC}{CD} = \frac{AC}{AD}$,$\therefore △ ABC ∽ △ ACD$.
$\therefore OA = 3$,$OB = 4$,$OD = 4$,$CD = 2$,
$\therefore AD = 4 - 1 = 1$,
$\therefore AB = \sqrt{3^2 + 4^2} = 5$,$BC = \sqrt{4^2 + (4 - 2)^2} = 2\sqrt{5}$,
$AC = \sqrt{(3 - 4)^2 + 2^2} = \sqrt{5}$.
$\because \frac{AB}{AC} = \frac{5}{\sqrt{5}} = \sqrt{5}$,$\frac{BC}{CD} = \frac{2\sqrt{5}}{2} = \sqrt{5}$,$\frac{AC}{AD} = \frac{\sqrt{5}}{1} = \sqrt{5}$,
$\therefore \frac{AB}{AC} = \frac{BC}{CD} = \frac{AC}{AD}$,$\therefore △ ABC ∽ △ ACD$.
5. 如图,在△ABC中,AB=25,BC=40,AC=20,在△ADE中,AE=12,AD=15,DE=24,求证:
(1)△ABC∽△ADE;
(2)△ADB∽△AEC.
(1)△ABC∽△ADE;
(2)△ADB∽△AEC.
答案
5 证明:
(1) $\because AB = 25$,$BC = 40$,$AC = 20$,$AE = 12$,$AD = 15$,$DE = 24$,
$\therefore \frac{AC}{AE} = \frac{AB}{AD} = \frac{BC}{DE} = \frac{5}{3}$,$\therefore △ ABC ∽ △ ADE$.
(2) $\because △ ABC ∽ △ ADE$,$\therefore ∠ BAC = ∠ DAE$,
$\therefore ∠ DAE - ∠ BAE = ∠ BAC - ∠ BAE$,$\therefore ∠ DAB = ∠ EAC$.
又$\because \frac{AD}{AE} = \frac{AB}{AC} = \frac{5}{4}$,$\therefore △ ADB ∽ △ AEC$.
(1) $\because AB = 25$,$BC = 40$,$AC = 20$,$AE = 12$,$AD = 15$,$DE = 24$,
$\therefore \frac{AC}{AE} = \frac{AB}{AD} = \frac{BC}{DE} = \frac{5}{3}$,$\therefore △ ABC ∽ △ ADE$.
(2) $\because △ ABC ∽ △ ADE$,$\therefore ∠ BAC = ∠ DAE$,
$\therefore ∠ DAE - ∠ BAE = ∠ BAC - ∠ BAE$,$\therefore ∠ DAB = ∠ EAC$.
又$\because \frac{AD}{AE} = \frac{AB}{AC} = \frac{5}{4}$,$\therefore △ ADB ∽ △ AEC$.
