《2026年山东省中考试卷精选九年级数学》

23. (本小题满分 11 分)[课本再现]
(1)如图(1)，在正方形 ABCD 中，对角线 AC 与 BD 相交于点 O，图中有8个等腰直角三角形.
[深入探究]
(2)如图(2)，在矩形 ABCD 中，AE，BE，CF，DF 分别为$\angle DAB$，$\angle ABC$，$\angle BCD$，$\angle ADC$的平分线，AE 与 DF 交于点 M，BE 与 CF 交于点 N.
①求证：四边形 EMFN 是正方形；
②连接 EF，求证：$EF = AB - BC$.
[结论应用]
(3)如图(3)，在四边形 ABCD 中，AE，BE，CF，DF 分别为$\angle DAB$，$\angle ABC$，$\angle BCD$，$\angle ADC$的平分线，AE 与 DF 交于点 M，BE 与 CF 交于点 N.若四边形 EMFN 是矩形，试判断四边形 ABCD 的形状，并给出证明.

答案：
23.[解析]本题考查了正方形的性质、解直角三角形、全等三角形的判定和性质.
(1)解:8
(2)①证明:$\because$四边形$ABCD$是矩形,$\therefore$四个内角均为$90°$.又$AE,BE,CF,DF$分别为$\angle DAB,\angle ABC,\angle BCD,\angle ADC$的平分线,$\therefore \angle EAB = \angle EBA = 45°$,$\therefore \triangle EBA$为等腰直角三角形$\therefore AE = BE$,$\angle E = 90°$.同理,$\angle F = \angle EMF = \angle ENF = 90°$,$\therefore$四边形$MFNE$为矩形.$\because AD = BC$,$\angle AMD = \angle BNC = 90°$,$\angle DAE = \angle CBN = 45°$,$\therefore \triangle DAM \cong \triangle CBN$,$\therefore AM = BN$,$\therefore ME = NE$,$\therefore$四边形$MFNE$是正方形.
②证明:如图,连接$EF$,由①知四边形$MFNE$是正方形,$\therefore \angle EFN = 45°$.$\because CF$平分$\angle BCD$,$\therefore \angle BCF = 45°$,$\therefore \angle EFN = \angle BCF$,$\therefore EF // BC$,$\therefore \frac{EF}{BC} = \frac{EN}{BN}$.$\because \triangle EBA$和$\triangle BCN$是等腰直角三角形.$\therefore BE = \frac{\sqrt{2}}{2} AB$,$BN = \frac{\sqrt{2}}{2} BC$,$\therefore EN = BE - BN = \frac{\sqrt{2}}{2} AB - \frac{\sqrt{2}}{2} BC = \frac{\sqrt{2}}{2} (AB - BC)$,$\therefore \frac{EF}{BC} = \frac{\frac{\sqrt{2}}{2} (AB - BC)}{\frac{\sqrt{2}}{2} BC} = \frac{AB - BC}{BC}$,$\therefore EF = AB - BC$.
(3)解:四边形$ABCD$是平行四边形.理由如下:$\because$四边形$EMFN$是矩形,$\therefore \angle E = 90°$,$\therefore \angle EAB + \angle EBA = 90°$.$\because \angle EMF = 90°$,$\therefore \angle DMA = 90°$.$\therefore \angle MAD + \angle MDA = 90°$.$\because AE$平分$\angle DAB$,$\therefore \angle EAB = \angle MAD$.$\because \angle EBA = \angle MDA$.$\because DF,BE$分别平分$\angle ADC,\angle ABC$,$\therefore \angle ADC = \angle ABC$.同理可证$\angle DAB = \angle DCB$,$\therefore \angle DAB + \angle ABC = \angle ADC + \angle DCB$.$\because \angle DAB + \angle ABC + \angle ADC + \angle DCB = 360°$,$\therefore \angle DAB + \angle ABC = 180°$,$\therefore AD // BC$.同理可证$DC // AB$,$\therefore$四边形$ABCD$是平行四边形.