《2025年配套综合练习甘肃八年级数学上册人教版》

如图 14.1 - 1,$\triangle AEC\cong\triangle ADB$.

(1) 写出它们的对应边和对应角；
(2) 若$\angle A = 50^{\circ}$,$\angle ABD = 39^{\circ}$,且$\angle 1 = \angle 2$,求$\angle 1$的度数.
【思路分析】(1) 根据$\triangle AEC\cong\triangle ADB$找出对应边和对应角,将其一一列出即可；(2) 根据全等三角形的性质结合三角形的内角和定理即可求出$\angle 1$的度数.

答案： 【解答】
(1)$\because\triangle AEC\cong\triangle ADB$,
$\therefore AE和AD$是对应边,$AC和AB$是对应边,$EC和DB$是对应边,$\angle AEC和\angle ADB$是对应角,$\angle ACE和\angle ABD$是对应角,$\angle A和\angle A$是对应角.
(2)$\because\triangle AEC\cong\triangle ADB$,$\therefore\angle ACE = \angle ABD = 39^{\circ}$.
$\because在\triangle ABC$中,$\angle A+\angle ABD+\angle 1+\angle 2+\angle ACE = 180^{\circ}$,$\angle A = 50^{\circ}$,$\angle ACE = \angle ABD = 39^{\circ}$,$\angle 1 = \angle 2$,$\therefore\angle 1 = 26^{\circ}$.