《2025年名校课堂八年级数学上册人教版》

6. 人大附中校本经典题 如图，$D$，$E$分别是$AB$，$AC$的中点，$CD \perp AB$，垂足为$D$，$BE \perp AC$，垂足为$E$，$CD$与$BE$相交于点$F$。
(1)求证：$AC = AB$。
(2)猜想$CF$与$DF$的数量关系，并证明。

答案： 解：
(1)证明：连接$BC.\because D$是$AB$的中点，$CD\perp AB$，$\therefore CA = CB$，同理$BA = BC$，$AC = AB$。
(2)猜想：$CF = 2DF$。证明：由
(1)得$AC = AB = BC$，$\therefore \triangle ABC$是等边三角形。$\therefore \angle A = \angle ABC = \angle ACB = 60^{\circ}$。在$Rt\triangle ABE$中，$\angle ABE = 90^{\circ} - \angle A = 30^{\circ}$，在$Rt\triangle BFD$中，$BF = 2DF$。在$Rt\triangle ADC$中，$\angle ACD = 90^{\circ} - \angle A = 30^{\circ}$，$\therefore \angle ABE = \angle ACD$。又$\because \angle ABC = \angle ACB$，$\therefore \angle FBC = \angle FCB$，$\therefore CF = BF$，$\therefore CF = 2DF$。

7. 北京四中校本经典题 如图，$\triangle ABC$为等腰三角形，$AC = BC$，$\triangle BDC$和$\triangle ACE$均为等边三角形，$AE$与$BD$相交于点$F$，连接$CF$交$AB$于点$G$。
(1)求证：$G$为$AB$的中点。
(2)若$\angle FAG = 15^{\circ}$，求$\angle BCE$的度数。

答案： 解：
(1)证明：$\because AC = BC$，$\angle CAB = \angle CBA$，$\because \triangle AEC$和$\triangle BCD$均为等边三角形，$\therefore \angle CAE = \angle CBD = 60^{\circ}$，$\therefore \angle CAE - \angle CAB = \angle CBD - \angle CBA$，即$\angle FAG = \angle FBG$，$\therefore AF = BF$。在$\triangle AFC$和$\triangle BFC$中，$\begin{cases}AF = BF\\AC = BC\\CF = CF\end{cases}$，$\therefore \triangle AFC\cong\triangle BFC(SSS).\therefore \angle ACF = \angle BCF$，即$CF$平分$\angle ACB$。又$\because AC = BC$，$\therefore AG = BG$，即$G$为$AB$的中点。
(2)设$BD$与$CE$相交于点$M$。由
(1)可得$\angle FBG = \angle FAG = 15^{\circ}$，$\therefore \angle BFE = \angle FBG + \angle FAG = 15^{\circ} + 15^{\circ} = 30^{\circ}$，$\therefore \angle E = 60^{\circ}$，$\therefore \angle EMF = 180^{\circ} - 30^{\circ} - 60^{\circ} = 90^{\circ}$。$\because CE\perp BD$。在$Rt\triangle BCM$中，$\angle BMC = 90^{\circ}$，$\angle CBD = 60^{\circ}$，$\therefore \angle BCE = 180^{\circ} - 90^{\circ} - 60^{\circ} = 30^{\circ}$。

8. 如图，在$\triangle ABC$中，$\angle BAC = 90^{\circ}$，$AB = AC$，$D$是$AB$的中点，连接$CD$，过点$B$作$BE \perp CD$交$CD$的延长线于点$E$，连接$AE$，过点$A$作$AF \perp AE$交$CD$于点$F$。求证：
(1)$AE = AF$。
(2)$CD = 2BE + DE$。

答案： 证明：
(1)$\because \angle BAC = 90^{\circ}$，$AF\perp AE$，$\therefore \angle EAB + \angle BAF = \angle BAF + \angle FAC = 90^{\circ}$，$\therefore \angle EAB = \angle FAC$。$\because BE\perp CD$，$\therefore \angle BEC = 90^{\circ}$，$\therefore \angle EBA + \angle EDB = \angle ADC + \angle FCA = 90^{\circ}$，$\therefore \angle EDB = \angle ADC$，$\therefore \angle EBA = \angle FCA$。在$\triangle AEB$和$\triangle AFC$中，$\begin{cases}\angle EAB = \angle FAC\\AB = AC\\\angle EBA = \angle FCA\end{cases}$，$\therefore \triangle AEB\cong\triangle AFC(ASA)$，$\therefore AE = AF$。
(2)过点$A$作$AG\perp EC$，垂足为$G.\because AG\perp EC$，$BE\perp CE$，$\therefore \angle BED = \angle AGD = 90^{\circ}$。$\because D$是$AB$的中点，$\therefore BD = AD$。在$\triangle BED$和$\triangle AGD$中，$\begin{cases}\angle BED = \angle AGD\\\angle EBD = \angle GAD\\BD = AD\end{cases}$，$\therefore \triangle BED\cong\triangle AGD(AAS)$，$\therefore ED = GD$，$BE = AG$。$\because AE = AF$，$AF\perp AF$，$\therefore \angle AEF = \angle AFE = 45^{\circ}$，$\therefore \angle FAG = 45^{\circ}$，$\therefore \angle GAF = \angle GFA$，$\therefore GA = GF$，$\therefore CF = BE = AG = GF$，$\therefore CD = DG + GF + FC = DE + BE + BE = 2BE + DE$。