《2025年假期面对面南方出版社八年级数学沪科版》

10. (玉林中考)如图,$\angle AOB = 60^{\circ}$,$OA = OB$,动点$C$从点$O$出发,沿射线$OB$方向移动,以$AC$为边在右侧作等边$\triangle ACD$,连接$BD$,则$BD$所在直线与$OA$所在直线的位置关系是 (A)

A. 平行
B. 相交
C. 垂直
D. 平行、相交或垂直

答案： A

11. (滨州中考)在等腰$\triangle ABC$中,$AB = AC$,$\angle B = 50^{\circ}$,则$\angle A$的大小为______$80^{\circ}$.

答案： $80^{\circ}$

12. (湘潭中考)如图,点$P$是$\angle AOC$的角平分线上一点,$PD\perp OA$,垂足为点$D$,且$PD = 3$,点$M$是射线$OC$上一动点,则$PM$的最小值为______3.

答案： 3

13. (襄阳中考)如图,在$\triangle ABC$中,$AB = AD = DC$,$\angle BAD = 20^{\circ}$,则$\angle C = $______40$^{\circ}$.

答案： 40

14. (常州中考)如图,在$\triangle ABC$中,$BC$的垂直平分线分别交$BC$,$AB$于点$E$,$F$.若$\triangle AFC$是等边三角形,则$\angle B = $______30$^{\circ}$.

答案： 30

15. (南京中考)如图,线段$AB$,$BC$的垂直平分线$l_1$,$l_2$相交于点$O$,若$\angle 1 = 39^{\circ}$,则$\angle AOC = $______78$^{\circ}$.

答案： 78

16. (12分)(长春中考)如图,图1,图2均是$8× 8$的小正方形网格,每个小正方形的顶点称为格点.线段$OM$,$ON$的端点均在格点上.在图1,图2给定的网格中以$OM$,$ON$为邻边各画一个四边形,使第四个顶点在格点上.要求:
(1)所画的两个四边形均是轴对称图形;
(2)所画的两个四边形不全等.

答案：
解:如图1,图2所示,(答案不唯一,正确即可)

17. (14分)(无锡中考)如图,在$\triangle ABC$中,$AB = AC$,点$D$,$E$分别在$AB$,$AC$上,$BD = CE$,$BE$,$CD$相交于点$O$.求证:

(1)$\triangle DBC\cong\triangle ECB$;
证明:$\because AB = AC$,$\therefore \angle DBC = \angle ECB$. 在$\triangle DBC$和$\triangle ECB$中,$\because \left\{ \begin{array} { l } { BD = CE, } \\ { \angle DBC = \angle ECB, } \\ { BC = CB, } \end{array} \right.$$\therefore \triangle DBC \cong \triangle ECB$.
(2)$OB = OC$.
证明:由(1)知$\triangle DBC \cong \triangle ECB$,$\therefore \angle DCB = \angle EBC$,$\therefore OB = OC$.

答案： 证明:
(1)$\because AB = AC$,$\therefore \angle DBC = \angle ECB$. 在$\triangle DBC$和$\triangle ECB$中,$\because \left\{ \begin{array} { l } { BD = CE, } \\ { \angle DBC = \angle ECB, } \\ { BC = CB, } \end{array} \right.$$\therefore \triangle DBC \cong \triangle ECB$.
(2)由
(1)知$\triangle DBC \cong \triangle ECB$,$\therefore \angle DCB = \angle EBC$,$\therefore OB = OC$.

18. (14分)(重庆中考)如图,在$\triangle ABC$中,$AB = AC$,$D$是$BC$边上的中点,连接$AD$.$BE$平分$\angle ABC$交$AC$于点$E$,过点$E$作$EF// BC$交$AB$于点$F$.
(1)若$\angle C = 36^{\circ}$,求$\angle BAD$的度数;54°
(2)求证:$FB = FE$.

答案：
(1)解:$\because AB = AC$,$\therefore \angle ABC = \angle C$. 又$\because D$是$BC$的中点,$\therefore AD$平分$\angle BAC$,即$\angle BAD = \frac { 1 } { 2 } \angle BAC$. $\because \angle C = 36 ^ { \circ }$,$\therefore \angle BAC = 180 ^ { \circ } - 36 ^ { \circ } - 36 ^ { \circ } = 108 ^ { \circ }$. $\therefore \angle BAD = 54 ^ { \circ }$.
(2)证明:$\because BE$平分$\angle ABC$,$\therefore \angle FBE = \angle EBD$. $\because EF // BC$,$\therefore \angle FEB = \angle EBD$,$\therefore \angle FBE = \angle FEB$,$\therefore FB = FE$.