【ID】2550 【题型】解答题 【类型】期中考试 【来源】2022年八年级数学上册第一次月考试卷

(1) 求证: $\triangle A C D \cong \triangle B C E$;
(2) 求证: $H C$ 平分 $\angle A H E$;
(3) 求 $\angle C H E$ 的度数. (用含 $\alpha$ 的式子表示)

(1) 证明: $\because \angle A C B=\angle D C E=\alpha$,
$$\therefore \angle A C D=\angle B C E \text {, }$$

\begin{aligned} &\left\{\begin{array}{l} \mathrm{CA}=\mathrm{CB} \\ \angle \mathrm{ACD}=\angle \mathrm{BCE}, \\ \mathrm{CD}=\mathrm{CE} \end{array}\right. \\ &\therefore \triangle A C D \cong \triangle B C E(S A S) ; \end{aligned}
(2) 证明: 过点 $C$ 作 $C M \perp A D$ 于 $M, C N \perp B E$ 于 $N$,
$\because \triangle A C D \cong \triangle B C E$,
$\therefore \angle C A M=\angle C B N$,

$\left\{\begin{array}{l}\angle \mathrm{CAM}=\angle \mathrm{CBN} \\ \angle \mathrm{AMC}=\angle \mathrm{BNC}=90^{\circ} \\ \mathrm{AC}=\mathrm{BC}\end{array}\right.$
$\therefore \triangle A C M \cong \triangle B C N(A A S)$,
$\therefore C M=C N$,
$\therefore H C$ 平分 $\angle A H E$;
(3) $\because \triangle A C D \cong \triangle B C E$,
$\therefore \angle C A D=\angle C B E$,
$\therefore \angle A H B=\angle A C B=\alpha$,
$\therefore \angle A H E=180^{\circ}-\alpha$,
$\therefore \angle C H E=\frac{1}{2} \angle A H E=90^{\circ}-\frac{1}{2} \alpha$.