《2025年一遍过九年级初中数学上册北师大版》

1 [2023东营中考]如图,△ABC为等边三角形,点D,E分别在边BC,AB上,∠ADE= 60°.若BD= 4DC,DE= 2.4,则AD的长为 (3)

A.1.8
B.2.4
C.3
D.3.2

答案： C $\because \triangle ABC$ 是等边三角形，$\therefore BC = AC$，$\angle B = \angle C = 60^{\circ}$，$\therefore \angle CAD + \angle ADC = 120^{\circ}$。$\because \angle ADE = 60^{\circ}$，$\therefore \angle BDE + \angle ADC = 120^{\circ}$，$\therefore \angle CAD = \angle BDE$，$\therefore \triangle ADC \backsim \triangle DEB$，$\therefore \frac{AD}{DE} = \frac{AC}{DB}$。由 $BD = 4DC$，可设 $DC = x$，则 $BD = 4x$，$\therefore BC = AC = 5x$，$\therefore \frac{AD}{2.4} = \frac{5x}{4x}$，$\therefore AD = 3$。

2 如图,在Rt△ABC中,∠ACB= 90°,D是边AB上一点,且CD= CA,BE⊥CD,交CD的延长线于点E.若$\frac{CD}{CE}= \frac{3}{5},BC= 9,$则BE的长为______6.

答案： 6 $\because \frac{CD}{CE} = \frac{3}{5}$，$\therefore$ 设 $CD = 3x$，$CE = 5x$，$\therefore DE = 2x$，$\therefore CA = CD = 3x$。$\because CD = CA$，$\therefore \angle A = \angle ADC$，$\because \angle ADC = \angle EDB$，$\therefore \angle A = \angle EDB$，$\because BE \perp CD$，$\therefore \angle ACB = \angle E = 90^{\circ}$，$\therefore \triangle ACB \backsim \triangle DEB$。$\therefore \frac{AC}{DE} = \frac{BC}{BE}$，$\therefore \frac{3x}{2x} = \frac{9}{BE}$，$\therefore BE = 6$。

3 [2024南京育英外国语学校月考]如图,在△ABC中,点D,E,G分别在边AB,AC,BC上,∠AED= ∠B,AG交线段DE于点F,且$\frac{AD}{AC}= \frac{DF}{CG}.(1)$求证:AG平分∠BAC.
证明：(1)∵∠DAE + ∠AED + ∠ADE = 180°，∠BAC + ∠B + ∠C = 180°，∠AED = ∠B，∴∠ADE = ∠C，又 ∵AD/AC = DF/CG，∴△ADF∽△ACG，∴∠DAF = ∠CAG，∴AG平分∠BAC。
(2)求证$\frac{EF}{BG}= \frac{DF}{CG}.$
证明：(2)∵∠AED = ∠B，∠EAF = ∠BAG，∴△AEF∽△ABG，∴EF/BG = AF/AG。由(1)知△ADF∽△ACG，∴DF/CG = AF/AG，∴EF/BG = DF/CG。

答案： 证明：
(1)$\because \angle DAE + \angle AED + \angle ADE = 180^{\circ}$，$\angle BAC + \angle B + \angle C = 180^{\circ}$，$\angle AED = \angle B$，$\therefore \angle ADE = \angle C$，又 $\because \frac{AD}{AC} = \frac{DF}{CG}$，$\therefore \triangle ADF \backsim \triangle ACG$，$\therefore \angle DAF = \angle CAG$，$\therefore AG$ 平分 $\angle BAC$。
(2)$\because \angle AED = \angle B$，$\angle EAF = \angle BAG$，$\therefore \triangle AEF \backsim \triangle ABG$，$\therefore \frac{EF}{BG} = \frac{AF}{AG}$。由
(1)知 $\triangle ADF \backsim \triangle ACG$，$\therefore \frac{DF}{CG} = \frac{AF}{AG}$，$\therefore \frac{EF}{BG} = \frac{DF}{CG}$。

4 如图,在菱形ABCD中,AB= AC,点E,F分别在边AB,BC上,且AE= BF,CE与AF相交于点G.
(1)求证:∠FGC= ∠B.
证明：∵ 四边形 ABCD 为菱形，∴ AB = BC，又 ∵ AB = AC，∴ AB = BC = AC，∴ △ABC 为等边三角形，∴ ∠B = ∠BAC = 60°。在 △ABF 和 △CAE 中，$\left\{\begin{array}{l} AB = CA, \\ \angle B = \angle CAE, \\ BF = AE, \end{array}\right.$ ∴ △ABF ≌ △CAE，∴ ∠BAF = ∠ACE，∴ ∠FGC = ∠GAC + ∠ACG = ∠GAC + ∠BAF = ∠BAC = 60°，∴ ∠FGC = ∠B。
(2)延长CE与DA的延长线交于点H,求证:BE·CH= AF·AC.
证明：∵ 四边形 ABCD 为菱形，∴ ∠B = ∠D，AD // BC，∴ ∠BCE = ∠H，∴ △BCE ∽ △DHC，∴ $\frac{BE}{DC} = \frac{CE}{HC}$。由(1)知 △ABF ≌ △CAE，∴ CE = AF。∵ CA = CB = CD，∴ $\frac{BE}{AC} = \frac{AF}{HC}$，∴ BE·CH = AF·AC。

答案： 证明：
(1)$\because$ 四边形 $ABCD$ 为菱形，$\therefore AB = BC$，又 $\because AB = AC$，$\therefore AB = BC = AC$，$\therefore \triangle ABC$ 为等边三角形，$\therefore \angle B = \angle BAC = 60^{\circ}$。在 $\triangle ABF$ 和 $\triangle CAE$ 中，$\left\{\begin{array}{l} AB = CA, \\ \angle B = \angle CAE, \\ BF = AE, \end{array}\right.$ $\therefore \triangle ABF \cong \triangle CAE$，$\therefore \angle BAF = \angle ACE$，$\therefore \angle FGC = \angle GAC + \angle ACG = \angle GAC + \angle BAF = \angle BAC = 60^{\circ}$，$\therefore \angle FGC = \angle B$。
(2)$\because$ 四边形 $ABCD$ 为菱形，$\therefore \angle B = \angle D$，$AD // BC$，$\therefore \angle BCE = \angle H$，$\therefore \triangle BCE \backsim \triangle DHC$，$\therefore \frac{BE}{DC} = \frac{CE}{HC}$。由
(1)知 $\triangle ABF \cong \triangle CAE$，$\therefore CE = AF$。$\because CA = CB = CD$，$\therefore \frac{BE}{AC} = \frac{AF}{HC}$，$\therefore BE \cdot CH = AF \cdot AC$。