《2025年名校课堂八年级数学上册沪科版》

1. 如图，在四边形 $ABCD$ 中，$AD // BC$，$\angle A = 90^{\circ}$，$BD = BC$，$CE \perp BD$ 于点 $E$。求证：$AD = BE$。

答案： 证明：$\because AD// BC$，$\therefore \angle ADB = \angle DBC$。又$\because CE\perp BD$，$\therefore \angle BEC = 90^{\circ}$。$\because \angle A = 90^{\circ}$，$\therefore \angle A = \angle BEC$。又$\because BD = CB$，$\therefore \triangle ABD \cong \triangle ECB(AAS)$。$\therefore AD = EB$。

2. 如图，$AD$ 是 $\triangle ABC$ 的中线，$BE \perp AD$，垂足为 $E$，$CF \perp AD$，交 $AD$ 的延长线于点 $F$，$G$ 是 $DA$ 延长线上一点，连接 $BG$。
(1) 求证：$BE = CF$；
(2) 若 $BG = CA$，求证：$GA = 2DE$。

答案：
(1)$\because AD$是$\triangle ABC$的中线，$\therefore BD = CD$。$\because BE\perp AD$，$CF\perp AD$，$\therefore \angle BED = \angle F = 90^{\circ}$。在$\triangle BED$和$\triangle CFD$中，$\begin{cases} \angle BED = \angle F \\ \angle BDE = \angle CDF \\ BD = CD \end{cases}$，$\therefore \triangle BED \cong \triangle CFD(AAS)$。$\therefore BE = CF$。
(2)在$Rt\triangle BGE$和$Rt\triangle CAF$中，$\begin{cases} BG = CA \\ BE = CF \end{cases}$，$\therefore Rt\triangle BGE \cong Rt\triangle CAF(HL)$。$\therefore GE = AF$。$\therefore AG = EF$。$\because \triangle BED \cong \triangle CFD$，$\therefore DE = DF$。$\therefore GA = 2DE$。

3. 如图，在 $\triangle ABC$ 中，过点 $C$ 作 $\angle BAC$ 的平分线 $AD$ 的垂线，垂足为 $D$，作 $DE // AB$，交 $AC$ 于点 $E$。求证：$AE = CE$。

答案： 证明：$\because AD$是$\angle BAC$的平分线，$\therefore \angle BAD = \angle DAE$。$\because DE// AB$，$\therefore \angle ADE = \angle BAD$。$\therefore \angle DAE = \angle ADE$。$\therefore AE = DE$。$\because AD\perp CD$，$\therefore \angle CAD + \angle ACD = 90^{\circ}$，$\angle ADE + \angle EDC = 90^{\circ}$。$\because \angle EDA = \angle EAD$，$\therefore \angle EDC = \angle ACD$。$\therefore DE = CE$。$\therefore AE = CE$。

4. (2024·安庆期末) 如图，$CD$ 为 $Rt\triangle ABC$ 斜边上的高，$\angle BAC$ 的平分线分别交 $CD$，$BC$ 于点 $E$，$F$，$FG \perp AB$，垂足为 $G$。
(1) 求证：$CE = FG$；
(2) 若 $AC = 12$，$AB = 15$，$CE = 4$，求 $\triangle ABC$ 的面积。

答案：
(1)证明：$\because AF$是$\angle BAC$的平分线，$\angle ACB = 90^{\circ}$，$FG\perp AB$，$CD\perp AB$，$\therefore FC = FG$，$\angle CAF = \angle DAE = \frac{1}{2}\angle BAC$，$\angle CAF + \angle CFA = 90^{\circ}$，$\angle DAE + \angle AED = 90^{\circ}$。$\therefore \angle AED = \angle AFC$。$\because \angle AED = \angle CEF$，$\therefore \angle CEF = \angle AFC$。$\therefore CE = CF$。$\therefore CE = FG$。
(2)$\because CE = 4$，$\therefore FG = CF = CE = 4$。$\because AC = 12$，$AB = 15$，$\therefore S_{\triangle ABC} = S_{\triangle ACF} + S_{\triangle ABF} = \frac{1}{2}AC\cdot CF + \frac{1}{2}AB\cdot FG = \frac{1}{2} × 12 × 4 + \frac{1}{2} × 15 × 4 = 54$。