初中数学八年级上册试题库（555题）

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如图2，当点F在AE的延长线上时，请猜想∠EFD与∠B，∠C之间的数量关系，并加以证明。

知识点：第十一章　三角形

参考答案：\(\angle EFD = \frac{1}{2}(\angle C - \angle B)\)（1）\(\because AE\)平分\(\angle BAC\)，\(\therefore \)\(\angle BAE = \frac{1}{2}\angle BAC = \frac{1}{2}(180^\circ - \angle B - \angle C) = 90^\circ - \frac{1}{2}(\angle B + \angle C)\)，\(\because \angle FEC = \angle B + \angle BAE\)，则\(\angle FEC = \angle B + 90^\circ - \frac{1}{2}(\angle B + \angle C) = 90^\circ + \frac{1}{2}(\angle B - \angle C)\)，\(\because FD \bot EC\)，\(\therefore \angle EFD = 90^\circ - \angle FEC\)，则\(\angle EFD = 90^\circ - [90^\circ + 2(\angle B - \angle C)] = \frac{1}{2}(\angle C - \angle B)\)，\(\because \angle B = 40^\circ \)，\(\angle C = 60^\circ \)，\(\therefore \angle EFD = \frac{1}{2}(60 - 40) = 10^\circ \)；（2）猜想：\(\angle EFD = \frac{1}{2}(\angle C - \angle B)\)．证明：同（1）可证：\(\angle AEC = 90^\circ + \frac{1}{2}(\angle B - \angle C)\)，\(\therefore \angle DEF = \angle AEC = 90^\circ + 2(\angle B - \angle C)\)，\(\therefore \)\(\angle EFD = 90^\circ - [90^\circ + \frac{1}{2}(\angle B - \angle C)] = \frac{1}{2}(\angle C - \angle B)\)．