《2025年自主学习指导课程与测试八年级数学上册人教版》

19. （8分）如图，点D，E分别是△ABC边AC，BC上的点，点D在线段AB的垂直平分线上，∠ABC=87°，∠ACB=33°，∠CAE=27°. 求证：△ABD是等边三角形.

答案： 19.证明:$\because \angle ACB = 33^{\circ}$,$\angle CAE = 27^{\circ}$,$\angle AEB$是$\triangle AEC$的外角,
$\therefore \angle AEB = \angle ACB + \angle CAE = 60^{\circ}$.
在$\triangle ABE$中,$\angle ABC = 87^{\circ}$,
$\therefore \angle BAE = 180^{\circ}-(\angle ABC + \angle AEB)=180^{\circ}-(87^{\circ}+60^{\circ}) = 33^{\circ}$.
$\therefore \angle BAC = \angle BAE + \angle CAE = 33^{\circ}+27^{\circ}=60^{\circ}$.
$\because$点$D$在线段$AB$的垂直平分线上,
$\therefore DA = DB$.
$\therefore \triangle ABD$是等边三角形.

20. （9分）如图，∠A=∠B，AE=BE，点D在AC边上，∠1=∠2，AE和BD相交于点O.
（1）求证：△AEC≌△BED.
（2）若∠1=46°，求∠BDE的度数.

答案： 20.
(1)证明:$\because AE$和$BD$相交于点$O$,$\therefore \angle AOD = \angle BOE$.
在$\triangle AOD$和$\triangle BOE$中,$\angle A = \angle B$,$\therefore \angle BEO = \angle2$.
又$\because \angle1 = \angle2$,$\therefore \angle1 = \angle BEO$.$\therefore \angle AEC = \angle BED$.
又$\because AE = BE$,
$\therefore \triangle AEC\cong \triangle BED(ASA)$.
(2)解:$\because \triangle AEC\cong \triangle BED$,$\therefore DE = CE$.$\therefore \angle EDC = \angle C$.
$\because \angle1 = 46^{\circ}$,$\therefore \angle EDC = \angle C = 67^{\circ}$.$\therefore \triangle AEC\cong \triangle BED$.
$\therefore \angle BDE = \angle C = 67^{\circ}$.