22. （9分）如图，在△ABC中，AB = 2BC，D、E分别是AB、AC的中点，将△ADE绕点E旋转180°得到△CFE. 试判断四边形BCFD的形状，并说明理由.
　　　

四边形$BCFD$是菱形, 理由是: $\because D, E$分别是$AB, AC$的中点, $\therefore 2DE = BC, DE// BC$. 又$\because \triangle CFE$是由$\triangle ADE$旋转而得, $\therefore DE = EF, \therefore DF = BC, DF// BC, \therefore$ 四边形$BCFD$是平行四边形. 又$\because AB = 2BC$, 且点$D$是$AB$的中点, $\therefore BD = BC, \therefore$ 四边形$BCFD$是菱形

23. （10分）已知：如图①，在正方形ABCD中，M是AB的中点，E是AB延长线上一点，MN⊥DM，且交∠CBE的平分线于点N.
（1）求证：DM = MN.
（2）若将上述条件“M是AB的中点”改为“M是AB上任意一点（不与点A、B重合）”，其余条件不变，如图②，DM与MN仍相等吗？为什么？

(1) 取$AD$的中点$F$, 连接$FM$. $\because \angle A = 90^{\circ}, \therefore \angle ADM + \angle AMD = 90^{\circ}$. $\because MN\perp DM, \therefore \angle AMD + \angle BMN = 90^{\circ}, \therefore \angle ADM = \angle BMN$①. $\because$ 四边形$ABCD$是正方形, $M, F$分别是$AB, AD$的中点, $\therefore DF = AF = AM = BM$. $\because \angle A = 90^{\circ}, \therefore \angle AFM = \angle AMF = 45^{\circ}, \angle DFM = 135^{\circ}$. $\because BN$是$\angle CBE$的平分线, $\therefore \angle CBN = 45^{\circ}, \angle DFM = \angle MBN = 135^{\circ}$②. $\because DF = BM$③, $\therefore \triangle DFM\cong \triangle MBN(ASA), \therefore DM = MN$; (2) 结论“$DM = MN$”仍成立. 证明: 在$AD$上截取$AF' = AM$, 连接$F'M$. $\because DF' = AD - AF', BM = AB - AM, AD = AB, AF' = AM, \therefore DF' = BM$. $\because \angle F'DM + \angle DMA = \angle BMN + \angle DMA = 90^{\circ}, \therefore \angle F'DM = \angle BMN$. 又$\angle DF'M = \angle MBN = 135^{\circ}$, 在$\triangle DF'M$和$\triangle MBN$中, $\angle F'DM = \angle BMN, DF' = BM, \angle DF'M = \angle MBN, \therefore \triangle DF'M\cong \triangle MBN, \therefore DM = MN$