《2025年安徽阳光夺冠单元与期末真题精选大试卷八年级数学上册人教版》

19. 如图，△ABC 中，AD 平分∠BAC，P 为 AD 延长线上一点，PE⊥BC 于 E，已知∠ACB = 80°，∠B = 24°，
（1）∠BAC 的度数为76°；
（2）求∠P 的度数.
解：∵AD平分∠BAC，∴∠BAD = $\frac{1}{2}$∠BAC = $\frac{1}{2}$×76° = 38°，∵∠BDP = ∠B + ∠BAD，∴∠BDP = 24° + 38° = 62°，∵PE⊥BC于E，∴∠PED = 90°，∴∠P = 90° - ∠BDP = 90° - 62° = 28°。即∠P的度数为28°。

答案： 19.解：
(1)76°；[解析] $\angle BAC = 180° - \angle B - \angle ACB = 76°$.
(2) $\because AD$平分$\angle BAC$，$\therefore \angle BAD = \frac{1}{2} \angle BAC = \frac{1}{2} × 76° = 38°$，$\because \angle BDP = \angle B + \angle BAD$，$\therefore \angle BDP = 24° + 38° = 62°$，$\because PE \perp BC$于$E$，$\therefore \angle PED = 90°$，$\therefore \angle P = 90° - \angle BDP = 90° - 62° = 28°$.即$\angle P$的度数为28°.

20. 如图，已知 DE // AC，∠A = ∠DEF.
（1）证明：EF // AB；
（2）各位同学，小学已经学习过“三角形的内角和为 180°”，利用图形，试证明此定理，即证明：∠A + ∠B + ∠C = 180°.

答案： 20.证明：
(1) $\because DE // AC$，$\therefore \angle A = \angle BDE$，$\because \angle A = \angle DEF$，$\therefore \angle BDE = \angle DEF$，$\therefore EF // AB$；
(2) $\because DE // AC$，$EF // AB$，$\therefore \angle C = \angle DEB$，$\angle B = \angle CEF$，$\because \angle DEB + \angle DEF + \angle CEF = 180°$，$\angle A = \angle DEF$，$\therefore \angle A + \angle B + \angle C = 180°$.

21. 已知：如图 1，在△ABC 中，CD 是高，若∠A = ∠DCB.
（1）试判断△ABC 的形状，并说明理由；
（2）如图 2，若 AE 是△ABC 的角平分线，AE，CD 相交于点 F. 求证：∠CFE = ∠CEF.

答案：
21.解：
(1) $\triangle ABC$是直角三角形.理由如下：$\because$在$\triangle ABC$中，$CD$是高，$\therefore \angle CDA = 90°$，$\therefore \angle A + \angle ACD = 90°$，$\because \angle A = \angle DCB$，$\therefore \angle DCB + \angle ACD = 90°$，$\because \angle ACB = 90°$，$\therefore \triangle ABC$是直角三角形；
(2)证明：$\because AE$是角平分线，$\therefore \angle DAF = \angle CAE$，$\because \angle FDA = 90°$，$\angle ACE = 90°$，$\therefore \angle DAF + \angle AFD = 90°$，$\angle CAE + \angle CEA = 90°$，$\therefore \angle AFD = \angle CEA$，$\because \angle AFD = \angle CFE$，$\therefore \angle CFE = \angle CEA$，即$\angle CFE = \angle CEF$.