《2025年1加1轻巧夺冠完美期末八年级数学上册人教版辽宁专版》

20. (8分)(抚顺望花期末)
如图,CE是$\triangle ABC$的外角$∠ACD$的平分线,且CE交BA的延长线于点E.
(1)求证:$∠BAC = ∠B + 2∠E$.
(2)若$∠B = 30^{\circ}$,$∠ACB = 40^{\circ}$,求$∠E$的度数.

答案： 20.
(1)证明:$\because CE$平分$\angle ACD$，$\therefore \angle ECD = \angle ACE$，$\because \angle BAC = \angle E + \angle ACE$，$\therefore \angle BAC = \angle E + \angle ECD$，$\because \angle ECD = \angle B + \angle E$，$\therefore \angle BAC = \angle E + \angle B + \angle E$，$\therefore \angle BAC = \angle B + 2\angle E$。
(2)解:$\because CE$平分$\angle ACD$，$\therefore \angle ACE = \angle ECD$，$\because \angle ACB = 40^{\circ}$，$\therefore \angle ACE = \angle ECD = \frac{1}{2}(180^{\circ} - 40^{\circ}) = 70^{\circ}$，$\therefore \angle BCE = \angle ACB + \angle ECA = 110^{\circ}$，$\therefore \angle E = 180^{\circ} - 30^{\circ} - 110^{\circ} = 40^{\circ}$。