《2025年广东名师讲练通九年级数学全一册北师大版深圳专版》

1. 如图,$AB$与$CD$相交于点$E$,$AD// BC,\frac {BE}{AE}=\frac {3}{5},CD=16$,则$DE$的长为10.

答案： $10$

2. 如图,$AC$是矩形$ABCD$的对角线,$E$是边$BC$延长线上一点,$AE$与$CD$相交于点$F$,则图中的相似三角形共有4对.

答案： $4$

3. (2024秋·宝安区校级月考)如图,下列条件不能判定$\triangle ADB\backsim \triangle ABC$的是(D).

A. $\angle ABD=\angle ACB$
B. $\angle ADB=\angle ABC$
C. $AB^{2}=AD\cdot AC$
D. $\frac {AD}{AB}=\frac {AB}{BC}$

答案： $D$

4. 如图,在$\triangle ABC$中,点$D$,$E$分别在$AB$,$AC$边上,$DE// BC$,且$\angle DCE=\angle B$,那么下列判断中,错误的是(D).

A. $\triangle ADE\backsim \triangle ABC$
B. $\triangle ADE\backsim \triangle ACD$
C. $\triangle DEC\backsim \triangle CDB$
D. $\triangle ADE\backsim \triangle DCB$

答案： $D$

5. 如图,在$\triangle ABC$中,点$D$在边$AB$上,满足$\angle ACD=\angle ABC$,若$AC=\sqrt {3},AD=1$,求$DB$的长.

解：$\because \angle A C D = \angle A B C$，$\angle A = \angle A$，
$\therefore \triangle A B C \backsim \triangle A C D$，$\therefore \frac { A D } { A C } = \frac { A C } { A B }$。
$\because A C = \sqrt { 3 }$，$A D = 1$，$\therefore \frac { 1 } { \sqrt { 3 } } = \frac { \sqrt { 3 } } { A B }$，
$\therefore A B = 3$，$\therefore B D = A B - A D = 3 - 1 =$2。

答案： 解：$\because \angle A C D = \angle A B C$，$\angle A = \angle A$，
$\therefore \triangle A B C \backsim \triangle A C D$，$\therefore \frac { A D } { A C } = \frac { A C } { A B }$。
$\because A C = \sqrt { 3 }$，$A D = 1$，$\therefore \frac { 1 } { \sqrt { 3 } } = \frac { \sqrt { 3 } } { A B }$，
$\therefore A B = 3$，$\therefore B D = A B - A D = 3 - 1 = 2$。

6. 如图,在$\triangle ABC$和$\triangle ADE$中,$\frac {AB}{AD}=\frac {BC}{DE}=\frac {AC}{AE}$,点$B$,$D$,$E$在一条直线上,求证:$\triangle ABD\backsim \triangle ACE$.
证明：$\because$在$\triangle A B C$和$\triangle A D E$中，$\frac { A B } { A D } = \frac { B C } { D E } = \frac { A C } { A E }$，$\therefore \triangle A B C \backsim \triangle A D E$，$\therefore \angle B A C = \angle D A E$，$\therefore \angle B A D = \angle C A E$。$\because \frac { A B } { A D } = \frac { A C } { A E }$，$\therefore \frac { A B } { A C } = \frac { A D } { A E }$，$\therefore \triangle A B D \backsim \triangle A C E$。

答案： 证明：$\because$在$\triangle A B C$和$\triangle A D E$中，$\frac { A B } { A D } = \frac { B C } { D E } = \frac { A C } { A E }$，
$\therefore \triangle A B C \backsim \triangle A D E$，
$\therefore \angle B A C = \angle D A E$，$\therefore \angle B A D = \angle C A E$。
$\because \frac { A B } { A D } = \frac { A C } { A E }$，$\therefore \frac { A B } { A C } = \frac { A D } { A E }$，$\therefore \triangle A B D \backsim \triangle A C E$。

7. (2024秋·龙华区期中)如图,矩形$ABCD$中,$E$为$BC$上一点,$DF\perp AE$于点$F$.
(1)证明$\triangle ABE\backsim \triangle DFA$;
(2)若$AB=3,AD=6,BE=4$,求$DF$的长.

(1)证明：$\because$四边形$A B C D$是矩形，
$\therefore A D // B C$，$\angle B = 90 ^ { \circ }$，$\therefore \angle A E B = \angle D A E$。
$\because D F \perp A E$，$\therefore \angle D A F + \angle A D F = \angle D A F + \angle E A B = 90 ^ { \circ }$，
$\therefore \angle A D F = \angle E A B$，$\therefore \triangle A B E \backsim \triangle D F A$。
(2)解：$\because A B = 3$，$B E = 4$，$\therefore$由勾股定理得$A E = 5$。
$\because \triangle A B E \backsim \triangle D F A$，$\therefore \frac { A E } { A B } = \frac { A D } { D F }$，即$\frac { 5 } { 3 } = \frac { 6 } { D F }$，$\therefore DF=$3.6。

答案： （1）证明：$\because$四边形$A B C D$是矩形，
$\therefore A D // B C$，$\angle B = 90 ^ { \circ }$，$\therefore \angle A E B = \angle D A E$。
$\because D F \perp A E$，$\therefore \angle D A F + \angle A D F = \angle D A F + \angle E A B = 90 ^ { \circ }$，
$\therefore \angle A D F = \angle E A B$，$\therefore \triangle A B E \backsim \triangle D F A$。
（2）解：$\because A B = 3$，$B E = 4$，$\therefore$由勾股定理得$A E = 5$。
$\because \triangle A B E \backsim \triangle D F A$，$\therefore \frac { A E } { A B } = \frac { A D } { D F }$，即$\frac { 5 } { 3 } = \frac { 6 } { D F }$，$\therefore D F = 3.6$。