“微信扫一扫”进入考试题库练习及模拟考试
初中数学八年级上册试题库(555题)
A.只有①③
B.只有①③④
C.只有②④
D.①②③④
参考答案：B
第48题
如图，在
参考答案：56°
第50题
求∠EAD的度数。
参考答案：20°（1）\(\because \angle B = 40^\circ \)，\(\angle ACB = 80^\circ \)，\(\therefore \angle BAC = 180^\circ - 40^\circ - 80^\circ = 60^\circ \)，\(\because AE\)是\(\Delta ABC\)的角平分线，\(\therefore \angle BAE = \frac{1}{2}\angle BAC = 30^\circ \)，\(\because FG \bot AE\)，\(\therefore \angle AHG = 90^\circ \)，\(\therefore \angle AGF = 180^\circ - 90^\circ - 30^\circ = 60^\circ \)；（2）\(\because AD\)是\(\Delta ABC\)的高，\(\therefore \angle ADC = 90^\circ \)，\(\because \angle ACB = 80^\circ \)，\(\therefore \angle CAD = 180^\circ - 90^\circ - 80^\circ = 10^\circ \)，\(\because \angle BAC = 60^\circ \)，\(AE\)是\(\Delta ABC\)的角平分线，\(\therefore \angle CAE = \frac{1}{2}\angle BAC = 30^\circ \)，\(\therefore \angle EAD = \angle CAE - \angle CAD = 30^\circ - 10^\circ = 20^\circ \)．
第52题
如图2，若∠ABC=α，∠BDA=β，求∠FAD+∠C的度数（用含α和β的代数式表示）。
参考答案：
β-0.2α
（1）
又
又
（2）如图2，过点
则
则
第54题
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。
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：\(\angle AEC = 90^\circ + \frac{1}{2}(\angle B - \angle C)\)，\(\therefore \angle DEF = \angle AEC = 90^\circ + 2(\angle B - \angle C)\)，\(\therefore \)\(\angle EFD = 90^\circ - [90^\circ + \frac{1}{2}(\angle B - \angle C)] = \frac{1}{2}(\angle C - \angle B)\)．
