《2025年全练单元卷九年级数学下册人教版》

26. （10分）如图，在矩形$ABCD$中，$E$为$AB$边上一点，$EC$平分$\angle DEB$，$F$为$CE$的中点，连接$AF$，$BF$，过点$E$作$EH// BC$分别交$AF$，$CD$于$G$，$H$两点。
（1）求证：$DE = DC$；
（2）求证：$AF\perp BF$；
（3）当$AF\cdot GF = 28$时，请直接写出$CE$的长.

答案：

(1)证明：$\because$四边形$ABCD$是矩形，$\therefore AB// CD$，$\therefore \angle DCE=\angle CEB$，$\because EC$平分$\angle DEB$，$\therefore \angle DEC=\angle CEB$，$\therefore \angle DCE=\angle DEC$，$\therefore DE = DC$；
(2)证明：连接$DF$， $\because DE = DC$，$F$为$CE$的中点，$\therefore DF\perp EC$，$\therefore \angle DFC = 90^{\circ}$，在矩形$ABCD$中，$AB = DC$，$\angle ABC = 90^{\circ}$，$\therefore BF = CF = EF=\frac{1}{2}EC$，$\therefore \angle ABF=\angle CEB$，$\because \angle DCE=\angle CEB$，$\therefore \angle ABF=\angle DCF$，在$\triangle ABF$和$\triangle DCF$中，$\begin{cases}BF = CF\\\angle ABF=\angle DCF\\AB = DC\end{cases}$，$\therefore \triangle ABF\cong\triangle DCF(SAS)$，$\therefore \angle AFB=\angle DFC = 90^{\circ}$，$\therefore AF\perp BF$；
(3)解：$CE = 4\sqrt{7}$.理由如下：$\because AF\perp BF$，$\therefore \angle BAF+\angle ABF = 90^{\circ}$，$\because EH// BC$，$\angle ABC = 90^{\circ}$，$\therefore \angle BEH = 90^{\circ}$，$\therefore \angle FEH+\angle CEB = 90^{\circ}$，$\because \angle ABF=\angle CEB$，$\therefore \angle BAF=\angle FEH$，$\because \angle EFG=\angle AFE$，$\therefore \triangle EFG\sim\triangle AFE$，$\therefore \frac{GF}{EF}=\frac{EF}{AF}$，即$EF^{2}=AF\cdot GF$，$\because AF\cdot GF = 28$，$\therefore EF = 2\sqrt{7}$，$\therefore CE = 2EF = 4\sqrt{7}$.