《2025年一遍过九年级初中数学上册北师大版》

1 [2024陕西师大附中期中]如图,添加以下哪个条件,仍不能直接证明△ABC与△ADE相似? (D)

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

答案： D A项，由两角分别相等的两个三角形相似，可判定$\triangle ABC$与$\triangle ADE$相似，A项不符合题意；B项，由两角分别相等的两个三角形相似，可判定$\triangle ABC$与$\triangle ADE$相似，B项不符合题意；C项，由两边成比例且夹角相等的两个三角形相似，可判定$\triangle ABC$与$\triangle ADE$相似，C项不符合题意。

2 [一题多解][2024贵港覃塘区期中]如图,在Rt△ABC中,∠ACB = 90°,AC = 6,BC = 12,点D在边BC上,点E在线段AD上,EF⊥AC于点F,EG⊥EF交AB于点G,若EF = EG,则CD的长为______.

答案：
4 解法一 如图1，过点D作$DM\perp BC$交AB于点M，易证$DC = DM$，设$CD = x$，则$DM = x$，由$\angle MDB=\angle ACB = 90^{\circ}$，$\angle B=\angle B$，得$\triangle BDM\backsim\triangle BCA$，所以$\frac{BD}{BC}=\frac{DM}{CA}$，即$\frac{12 - x}{12}=\frac{x}{6}$，解得$x = 4$。
相似三角形的常见模型
　　　　　　　　　　 　　　
解法二 如图2，延长FE，交AB于点N，易得$Rt\triangle ENG\backsim Rt\triangle CBA$，$\therefore EN:EG = CB:CA = 12:6 = 2:1$，又$\because EF = EG$，$\therefore EN:EF = 2:1$。由题中条件易得$\triangle AEN\backsim\triangle ADB$，$\triangle AEF\backsim\triangle ADC$，$\therefore EN:BD = AE:AD$，$EF:CD = AE:AD$，$\therefore EN:BD = EF:CD$，$\therefore BD:CD = EN:EF = 2:1$，$\therefore CD=\frac{1}{3}BC=\frac{1}{3}×12 = 4$。

3 [2025无锡期末]如图,在△ABC中,点D,E分别在BC,AC边上,且AD = AB,∠DEC = ∠B.
(1)求证:△AED∽△ADC.
证明：$\because AD = AB$，$\therefore\angle ADB=\angle B$，$\because\angle DEC=\angle B$，$\therefore\angle DEC=\angle ADB$，又$\because\angle DEC+\angle AED = 180^{\circ}$，$\angle ADB+\angle ADC = 180^{\circ}$，$\therefore\angle AED=\angle ADC$，又$\because\angle EAD=\angle DAC$，$\therefore\triangle AED\backsim\triangle ADC$。
(2)若AB = 2,EC = 3,求AC的长.
解：由(1)可知，$\triangle AED\backsim\triangle ADC$，$\therefore\frac{AE}{AD}=\frac{AD}{AC}$，$\because AB = AD = 2$，$CE = 3$，$\therefore\frac{AC - 3}{2}=\frac{2}{AC}$，解得$AC = 4$或$-1$（不合题意，舍去），$\therefore AC$的长为4。

答案：
(1)证明：$\because AD = AB$，$\therefore\angle ADB=\angle B$，$\because\angle DEC=\angle B$，$\therefore\angle DEC=\angle ADB$，又$\because\angle DEC+\angle AED = 180^{\circ}$，$\angle ADB+\angle ADC = 180^{\circ}$，$\therefore\angle AED=\angle ADC$，又$\because\angle EAD=\angle DAC$，$\therefore\triangle AED\backsim\triangle ADC$。
(2)解：由
(1)可知，$\triangle AED\backsim\triangle ADC$，$\therefore\frac{AE}{AD}=\frac{AD}{AC}$，$\because AB = AD = 2$，$CE = 3$，$\therefore\frac{AC - 3}{2}=\frac{2}{AC}$，解得$AC = 4$或$-1$（不合题意，舍去），$\therefore AC$的长为4。

4 如图,∠BAC = 90°,AD⊥BC于点D,AE = EC,ED的延长线交AB的延长线于点F.求证:$\frac{AB}{AC}= \frac{DF}{AF}$.

证明：$\because AD\perp BC$，$\therefore\angle ADC = 90^{\circ}$。$\because AE = CE$，$\therefore DE$是$Rt\triangle ACD$斜边上的中线，$\therefore DE = CE$，$\therefore\angle C=\angle EDC$，又$\because\angle BDF=\angle EDC$，$\therefore\angle BDF=\angle C$。$\because\angle BAD+\angle ABD = 90^{\circ}$，$\angle C+\angle ABD = 90^{\circ}$，$\therefore\angle BAD=\angle C$，$\therefore\angle BDF=\angle BAD$，又$\because\angle F=\angle F$，$\therefore\triangle DBF\backsim\triangle ADF$，$\therefore\frac{BD}{AD}=\frac{DF}{AF}$。$\because\angle ADB=\angle CDA = 90^{\circ}$，$\angle BAD=\angle C$，$\therefore\triangle ABD\backsim\triangle CAD$，$\therefore\frac{AB}{AC}=\frac{BD}{AD}$，$\therefore\frac{AB}{AC}=\frac{DF}{AF}$。

答案： 证明：$\because AD\perp BC$，$\therefore\angle ADC = 90^{\circ}$。$\because AE = CE$，$\therefore DE$是$Rt\triangle ACD$斜边上的中线，$\therefore DE = CE$，$\therefore\angle C=\angle EDC$，又$\because\angle BDF=\angle EDC$，$\therefore\angle BDF=\angle C$。$\because\angle BAD+\angle ABD = 90^{\circ}$，$\angle C+\angle ABD = 90^{\circ}$，$\therefore\angle BAD=\angle C$，$\therefore\angle BDF=\angle BAD$，又$\because\angle F=\angle F$，$\therefore\triangle DBF\backsim\triangle ADF$，$\therefore\frac{BD}{AD}=\frac{DF}{AF}$。$\because\angle ADB=\angle CDA = 90^{\circ}$，$\angle BAD=\angle C$，$\therefore\triangle ABD\backsim\triangle CAD$，$\therefore\frac{AB}{AC}=\frac{BD}{AD}$，$\therefore\frac{AB}{AC}=\frac{DF}{AF}$。