《2025年学霸题中题八年级数学上册人教版》

1. 如图,在$\triangle ABC$中,$AB = AC$,$AD \perp BC$,下列结论中不一定正确的是 (C)

A. D是BC的中点
B. AD平分$\angle BAC$
C. $AB = 2BD$
D. $\angle B = \angle C$

答案： C

2. 改编题 下列说法正确的有 (A)
①等腰三角形的高、中线、角平分线互相重合;
②等腰三角形的高一定平分对边;③等腰三角形的角平分线就是它对边的中线;④等腰三角形是轴对称图形,底边上的高是它的对称轴;
⑤等腰三角形的底边上的垂直平分线是它的对称轴.
A. 1个
B. 2个
C. 3个
D. 4个

答案： A

3. (2024·兰州中考)如图,在$\triangle ABC$中,$AB = AC$,$\angle BAC = 130^{\circ}$,$DA \perp AC$,则$\angle ADB =$ (B)
A. $100^{\circ}$
B. $115^{\circ}$
C. $130^{\circ}$
D. $145^{\circ}$

答案： B

4. (1)(2024·湖南中考)等腰三角形一个底角的度数是$40^{\circ}$,则其顶角的度数为$100^{\circ}$.
(2)已知等腰三角形的一个外角为$130^{\circ}$,则它的顶角的度数为$50^{\circ}$或$80^{\circ}$.

答案：
(1) $100^{\circ}$
(2) $50^{\circ}$或 $80^{\circ}$

5. (鞍山中考改编)如图,以$\triangle ABC$的顶点B为圆心,BA长为半径画弧,交BC边于点D,连接AD.若$\angle B = 40^{\circ}$,$\angle C = 36^{\circ}$,则$\angle DAC$的大小为34$^{\circ}$.

答案： 34

6. (2024·泰安月考)如图,在$\triangle ABC$中,$\angle BAC = 90^{\circ}$,E为边BC上的点,且$AB = AE$,D为线段BE的中点,过点E作$EF \perp AE$,过点A作$AF // BC$,且AF,EF相交于点F.求证:
(1)$\angle C = \angle BAD$;
证明:$ \because AB = AE $，D 为线段 BE 的中点，$ \therefore AD \perp BC $，$ \therefore \angle C + \angle DAC = 90^{\circ} $。$ \because \angle BAC = 90^{\circ} $，$ \therefore \angle BAD + \angle DAC = 90^{\circ} $，$ \therefore \angle C = \angle BAD $。
(2)$AC = EF$;
证明:$ \because AF // BC $，$ \therefore \angle FAE = \angle AEB $。$ \because AB = AE $，$ \therefore \angle B = \angle AEB $，$ \therefore \angle B = \angle FAE $。又 $ \angle AEF = \angle BAC = 90^{\circ} $，$ AB = AE $，$ \therefore \triangle ABC \cong \triangle EAF (ASA) $，$ \therefore AC = EF $。

答案：
(1) $ \because AB = AE $，D 为线段 BE 的中点，$ \therefore AD \perp BC $，$ \therefore \angle C + \angle DAC = 90^{\circ} $。$ \because \angle BAC = 90^{\circ} $，$ \therefore \angle BAD + \angle DAC = 90^{\circ} $，$ \therefore \angle C = \angle BAD $。
(2) $ \because AF // BC $，$ \therefore \angle FAE = \angle AEB $。$ \because AB = AE $，$ \therefore \angle B = \angle AEB $，$ \therefore \angle B = \angle FAE $。又 $ \angle AEF = \angle BAC = 90^{\circ} $，$ AB = AE $，$ \therefore \triangle ABC \cong \triangle EAF (ASA) $，$ \therefore AC = EF $。

7. (宁夏中考)如图,在$\triangle ABC$中,$AC = BC$,点D和E分别在AB和AC上,且$AD = AE$.连接DE,过点A的直线GH与DE平行,若$\angle C = 40^{\circ}$,则$\angle GAD$的度数为 (C)

A. $40^{\circ}$
B. $45^{\circ}$
C. $55^{\circ}$
D. $70^{\circ}$

答案： C 解析：$ \because AC = CB $，$ \angle C = 40^{\circ} $，$ \therefore \angle BAC = \angle B = \frac{1}{2}(180^{\circ} - 40^{\circ}) = 70^{\circ} $。$ \because AD = AE $，$ \therefore \angle ADE = \angle AED = \frac{1}{2}(180^{\circ} - 70^{\circ}) = 55^{\circ} $。$ \because GH // DE $，$ \therefore \angle GAD = \angle ADE = 55^{\circ} $，故选 C。

8. (衢州中考)“三等分角”大约是在公元前五世纪由古希腊人提出来的,借助如图①所示的“三等分角仪”能三等分任一角.如图②,这个三等分角仪由两根有槽的棒OA,OB组成,两根棒在O点相连并可绕O转动,C点固定,$OC = CD = DE$,点D,E可在槽中滑动.若$\angle BDE = 75^{\circ}$,则$\angle CDE$的度数是 (D)

A. $60^{\circ}$
B. $65^{\circ}$
C. $75^{\circ}$
D. $80^{\circ}$

答案： D 解析：$ \because OC = CD = DE $，$ \therefore \angle O = \angle ODC $，$ \angle DCE = \angle DEC $。$ \therefore \angle DCE = \angle O + \angle ODC = 2\angle ODC $。$ \because \angle O + \angle OED = 3\angle ODC = \angle BDE = 75^{\circ} $，$ \therefore \angle ODC = 25^{\circ} $。$ \because \angle CDE + \angle ODC = 180^{\circ} - \angle BDE = 105^{\circ} $，$ \therefore \angle CDE = 105^{\circ} - \angle ODC = 80^{\circ} $。