1. 在$△ ABC$和$△ A_{1}B_{1}C_{1}$中，$∠ C = ∠ C_{1} = 90°$，$AC = 12$，$AB = 15$，$A_{1}C_{1} = 8$，要使$△ ABC ∽ △ A_{1}B_{1}C_{1}$，则$A_{1}B_{1}$的长为（）

A.8
B.10
C.12
D.14

依题意,知$△ ABC$和$△ A_{1}B_{1}C_{1}$都是直角三角形,$AB$和$A_{1}B_{1}$是斜边,当$\frac{AB}{A_{1}B_{1}} = \frac{AC}{A_{1}C_{1}}$,即$\frac{15}{A_{1}B_{1}} = \frac{12}{8}$,$A_{1}B_{1} = 10$ 时,$△ ABC ∽ △ A_{1}B_{1}C_{1}$.

2. 教材P98例6变式 如图，$∠ ACB = ∠ ADC = 90°$，$AC = \sqrt{6}$，$AD = 2$. 当AB的长为多少时，$\mathrm{Rt}△ ACD$与$\mathrm{Rt}△ ABC$相似？

解:$\because AC = \sqrt{6} ,AD = 2,∠ ADC = 90°$,
$\therefore CD = \sqrt{AC^2 - AD^2} = \sqrt{2}$.
要使$\mathrm{Rt}△ ACD$与$\mathrm{Rt}△ ABC$相似,有两种情况:
①当$\frac{AC}{AD} = \frac{AB}{AC}$时,有$\mathrm{Rt}△ ABC ∽ \mathrm{Rt}△ ACD$,
$\therefore \frac{\sqrt{6}}{2} = \frac{AB}{\sqrt{6}} ,\therefore AB = 3$.
②当$\frac{AC}{CD} = \frac{AB}{CA}$时,有$\mathrm{Rt}△ ACB ∽ \mathrm{Rt}△ CDA$,
$\therefore \frac{\sqrt{6}}{\sqrt{2}} = \frac{AB}{\sqrt{6}} ,\therefore AB = 3\sqrt{2}$.
综上,当$AB$的长为$3$或$3\sqrt{2}$时,$\mathrm{Rt}△ ACD$与$\mathrm{Rt}△ ABC$相似.

3. [2026 亳州期中] 如图,已知在△ABC与△ADE中,∠C=∠AED=90°,点E在AB上,那么添加下列一个条件后,仍然不能判定△ABC与△DAE相似的是 ()

A.∠CAB=∠D
B.AD//BC
C.$\frac{AB}{AC}=\frac{AD}{DE}$
D.$\frac{BC}{AC}=\frac{AD}{AE}$

解题思路:已知$∠ C = ∠ AED = 90°$,添加另外两组角中的一组角对应相等或添加等角的两组边对应成比例,因为是直角三角形,所以还可添加斜边、直角边对应成比例.
A项,由$∠ C = ∠ AED = 90°$,$∠ CAB = ∠ D$,可知$△ ABC ∽ △ DAE$;B项,由$AD // BC$,可得$∠ B = ∠ DAE$,再结合$∠ C = ∠ AED = 90°$,可得$△ ABC ∽ △ DAE$;易知C项,可判定$△ ABC ∽ △ DAE$.

4. 如图,在 $Rt△ ABC$ 中, $∠ BAC = 90°$,AD 是斜边 BC 上的高,其中正确的是 ()
①$AB^2 = BD · BC$;②$AD^2 = BD · CD$;
③$AC^2 = CD · CB$;④$AB · AC = AD · CB$.

A.①②③
B.①②③④
C.①④
D.①③④

$\because AD$ 是斜边 $BC$ 上的高,$\therefore ∠ ADC = ∠ ADB = 90°$.
又$\because ∠ BAC = 90°$,$\therefore ∠ BAC = ∠ ADB = 90°$. 又$\because ∠ B = ∠ B$,
$\therefore △ BAD ∽ △ BCA$,$\therefore \frac{BD}{AB} = \frac{AB}{BC}$,即 $AB^2 = BD · BC$,故①正确.
$\because △ BAD ∽ △ BCA$,$\therefore ∠ C = ∠ BAD$. 又$\because ∠ ADC = ∠ ADB = 90°$,$\therefore △ BAD ∽ △ ACD$,$\therefore \frac{BD}{AD} = \frac{AD}{CD}$,即 $AD^2 = BD · CD$,故②正确. $\because ∠ BAC = ∠ ADC = 90°$,$∠ C = ∠ C$,$\therefore △ BAC ∽ △ ADC$,
$\therefore \frac{AC}{CD} = \frac{CB}{AC}$,即 $AC^2 = CD · CB$,故③正确. $\because S_{△ ABC} = \frac{1}{2} AB · AC = \frac{1}{2} BC · AD$(等面积法),$\therefore AB · AC = AD · CB$,故④正确. 综上可知,正确的是①②③④.

5. [2025连云港中考]如图,在$△ ABC$中,$∠ ACB = 90°$,$∠ CAB = 30°$,$AD$平分$∠ CAB$,$BE ⊥ AD$,$E$为垂足,则$\frac{AD}{BE}$的值为 ()


A.$2\sqrt{3}$
B.$\frac{7\sqrt{3}}{3}$
C.$\frac{5\sqrt{3}}{2}$
D.$\frac{8\sqrt{3}}{3}$

如图,延长 $BE$ 交 $AC$ 的延长线于点 $F$. $\because AD$ 平分$∠ CAB$,$\therefore ∠ FAE = ∠ BAE$. 又$\because AE ⊥ BF$,$\therefore ∠ ABF = ∠ F$,
$\therefore AF = AB$,$\therefore BE = EF$(点拨:见角平分线 + 垂直,考虑“三线合一”). $\because ∠ ACD = ∠ BED = 90°$,$∠ ADC = ∠ BDE$,$\therefore ∠ CAD = ∠ CBF$. 又$\because ∠ ACD = ∠ BCF$,$\therefore △ ACD ∽ △ BCF$,$\therefore \frac{AD}{BF} = \frac{AC}{BC} = \sqrt{3}$,$\therefore \frac{AD}{BE} = \frac{AD}{\frac{1}{2}BF} = \frac{2AD}{BF} = 2\sqrt{3}$. 故选A.

6. 教材P123T2变式 如图,在$Rt△ ACB$中,$∠ C=90°$,直角边AC上有一点D(不与点A,C重合),过点D作直线截$△ ACB$,使截得的三角形与$△ ACB$相似,则满足条件的直线共有条.

如图,①过点 $D$ 作 $DP ⊥ AB$,垂足为 $P$,则$△ APD ∽ △ ACB$;②过点 $D$ 作 $DE // BC$,交 $AB$ 于点 $E$,则$△ ADE ∽ △ ACB$;③过点 $D$ 作 $DF // AB$,交 $BC$ 于点 $F$,则$△ DCF ∽ △ ACB$;④在 $BC$ 上取一点 $G$,使得$∠ DGC = ∠ A$,则$△ GCD ∽ △ ACB$. 综上,满足条件的直线共有4条.

7. [2024 内蒙古呼伦贝尔中考]如图,点$A(0,-2)$,$B(1,0)$,将线段$AB$平移得到线段$DC$,若$∠ ABC=90°$,$BC=2AB$,则点$D$的坐标是.

如图,过点 $D$ 作 $DE ⊥ y$ 轴于点 $E$. $\because$ 点 $A(0,-2)$,$B(1,0)$,$\therefore OA = 2$,$OB = 1$. 由平移的性质,得 $AB // CD$,$AB = CD$,$\therefore$ 四边形 $ABCD$ 是平行四边形. 又$\because ∠ ABC = 90°$,
$\therefore$ 四边形 $ABCD$ 是矩形,$\therefore ∠ BAD = 90°$,$AD = BC = 2AB$,
$\therefore ∠ BAO + ∠ DAE = 90°$. 又$\because ∠ BAO + ∠ ABO = 90°$,$\therefore ∠ ABO = ∠ DAE$. 又$\because ∠ AOB = ∠ DEA = 90°$,$\therefore △ ABO ∽ △ DAE$(“一线三等角”模型总结见本章专项1类型3),$\therefore \frac{OA}{DE} = \frac{OB}{AE} = \frac{AB}{AD} = \frac{1}{2}$,$\therefore DE = 2OA = 4$,$AE = 2OB = 2$,$\therefore OE = OA + AE = 4$,
$\therefore D(4,-4)$.

8. [2026合肥瑶海区段考]如图,已知CD是Rt△ABC斜边AB上的中线,DE⊥AB交BC于点F,交AC的延长线于点E.
求证:(1)△ADE∽△FDB
(2)CD² = DE·DF.

证明:(1)$\because DE ⊥ AB$,$△ ABC$ 是直角三角形,
$\therefore ∠ ECB = ∠ EDB = 90°$.
$\because ∠ DFB = ∠ CFE$,$\therefore ∠ DEA = ∠ DBF$.
又$\because ∠ ADE = ∠ FDB = 90°$,$\therefore △ ADE ∽ △ FDB$.
(2)由(1)知$△ ADE ∽ △ FDB$,$\therefore \frac{DE}{DB} = \frac{DA}{DF}$.
$\because CD$ 是 $\mathrm{Rt}△ ABC$ 斜边 $AB$ 上的中线,
$\therefore DA = DB = CD$,$\therefore \frac{DE}{CD} = \frac{CD}{DF}$,
$\therefore CD^2 = DE · DF$.