《2025年新课程新教材导航学数学九年级上册北师大版》

14. （12 分）如图，在矩形 $ABCD$ 中，$E$ 是边 $AD$ 上的一个动点，点 $F,G,H$ 分别是边 $BC,BE,CE$ 的中点.
（1）求证：$\triangle BGF\cong\triangle FHC$.
（2）设 $AD = a$，当四边形 $EGFH$ 是正方形时，求矩形 $ABCD$ 的面积.

答案： 14.
(1)证明：$\because$点$F$，$G$，$H$分别是边$BC$，$BE$，$CE$的中点，$\therefore FH // BE$，$FH = \frac{1}{2}BE$，$BG = GE = \frac{1}{2}BE$，$BF = CF$.$\therefore \angle CFH = \angle CBG$，$FH = BG$.$\therefore \triangle BGF \cong \triangle FHC$.
(2)解：连接$EF$，$GH$.当四边形$EGFH$是正方形时，$EF \perp GH$且$EF = GH$.$\because$在$\triangle BEC$中，点$G$，$H$分别是$BE$，$CE$的中点，$\therefore GH = \frac{1}{2}BC = \frac{1}{2}AD = \frac{1}{2}a$，$GH // BC$，$\therefore EF \perp BC$，$\therefore \angle EFB = 90^{\circ}$.$\because$四边形$ABCD$是矩形，$\therefore \angle A = \angle ABC = 90^{\circ}$.$\therefore$四边形$AEFB$是矩形，$\therefore AB = EF = GH = \frac{1}{2}a$，$\therefore$矩形$ABCD$的面积为$AB · AD = \frac{1}{2}a · a = \frac{1}{2}a^{2}$.

15. （14 分）如图 1，点 $E$ 为正方形 $ABCD$ 的边 $AB$ 上一点，$EF\perp EC$，且 $EF = EC$，连接 $AF$.
（1）求 $\angle EAF$ 的度数.
（2）如图 2，连接 $FC$ 交 $BD$ 于点 $M$，交 $AD$ 于点 $N$. 求证：$BD = AF + 2DM$.

答案： 15.
(1)解：过点$F$作$FM \perp AB$交$BA$的延长线于点$M$.$\because$四边形$ABCD$是正方形，$EF \perp EC$，$\therefore \angle B = \angle M = \angle CEF = 90^{\circ}$，$\therefore \angle MEF + \angle CEB = 90^{\circ}$，$\angle CEB + \angle BCE = 90^{\circ}$，$\therefore \angle MEF = \angle BCE$.$\because EC = FE$，$\therefore \triangle EBC \cong \triangle FME$.$\therefore FM = EB$，$EM = BC$.$\because BC = AB$，$\therefore EM = AB$，$\therefore EM - AE = AB - AE$，$\therefore AM = EB$，$\therefore FM = AM$.$\because \angle M = 90^{\circ}$，$\therefore \angle MAF = 45^{\circ}$，$\therefore \angle EAF = 180^{\circ} - 45^{\circ} = 135^{\circ}$.
(2)证明：过点$F$作$FG // AB$交$BD$于点$G$.由
(1)可知$\angle EAF = 135^{\circ}$.$\because \angle ABD = 45^{\circ}$，$\therefore \angle EAF + \angle ABD = 135^{\circ} + 45^{\circ} = 180^{\circ}$，$\therefore AF // BG$.又$\because FG // AB$，$\therefore$四边形$ABGF$为平行四边形，$\therefore AF = BG$，$FG = AB$.$\because AB = CD$，$\therefore FG = CD$.$\because AB // CD$，$\therefore FG // CD$，$\therefore \angle FGM = \angle CDM$.又$\because \angle FMG = \angle CMD$，$\therefore \triangle FGM \cong \triangle CDM$，$\therefore GM = DM$，$\therefore DG = 2DM$.$\therefore BD = BG + DG = AF + 2DM$.