23. (10分)如图,四边形$ABCD$是正方形,点$E$,$F$分别在$BC$,$AB$上,点$M$在$BA$的延长线上,且$CE = BF = AM$,过点$M$,$E$分别作$DM$,$DE$的垂线,相交于点$N$,连接$NF$.
(1)求证:$DE ⊥ DM$;
(2)猜想四边形$CENF$是怎样的特殊四边形,并证明你的猜想.

23. 解:(1)证明:
$\because$四边形$ABCD$是正方形,
$\therefore DC = DA$,$∠ DCE = ∠ DAM = ∠ ADC = 90^{ \circ }$,
$\therefore △ DCE≌△ DAM(\mathrm{SAS})$,
$\therefore DE = DM$,$∠ EDC = ∠ MDA$.
又$\because ∠ ADE + ∠ EDC = ∠ ADC = 90^{ \circ }$,
$\therefore ∠ ADE + ∠ MDA = 90^{ \circ }$,$\therefore DE ⊥ DM$.
(2)解:四边形$CENF$是平行四边形.理由如下:
$\because$四边形$ABCD$是正方形,$\therefore AB // CD$,$AB = CD$.
$\because BF = AM$,$\therefore MF = AF + AM = AF + BF = AB$,即$MF = CD$.$\because$点$F$在$AB$上,点$M$在$BA$的延长线上,
$\therefore MF // CD$,$\therefore$四边形$CFMD$是平行四边形,
$\therefore DM = CF$,$DM // CF$.
$\because NM ⊥ DM$,$NE ⊥ DE$,$DE ⊥ DM$,
$\therefore$四边形$DENM$是矩形,$\therefore EN = DM$,$EN // DM$,
$\therefore CF = EN$,$CF // EN$,$\therefore$四边形$CENF$为平行四边形.

24. (12分)如图,在四边形$ABCD$中,$BA = BC$,$∠ ABC = 60°$,$∠ ADC = 30°$,连接对角线$BD$.
(1)将线段$CD$绕点$C$顺时针旋转$60°$得到线段$CE$,连接$AE$.
①依题意补全图.
②试判断$AE$与$BD$的数量关系,并证明你的结论.
(2)在(1)的条件下,直接写出线段$DA$,$DB$和$DC$之间的数量关系.
(3)若$F$是对角线$BD$上一点,且满足$∠ AFC = 150°$,连接$FA$和$FC$,探究线段$FA$,$FB$和$FC$之间的数量关系,并证明.

24. 解:(1)①补全图形如图①.

②$AE = BD$.证明:如图②,连接$AC$,
$\because BA = BC$,$∠ ABC = 60^{ \circ }$,
$\therefore △ ABC$是等边三角形,$\therefore ∠ ACB = 60^{ \circ }$,$CA = CB$,
$\because$将线段$CD$绕点$C$顺时针旋转$60^{ \circ }$得到线段$CE$,
$\therefore CD = CE$,$∠ DCE = 60^{ \circ }$,
$\therefore ∠ BCD = ∠ ACE$.
(2)$DA^{2}+DC^{2}=DB^{2}$.
(3)当点$F$在直线$AC$左侧时,$FA^{2}+FC^{2}=FB^{2}$;
当点$F$在直线$AC$右侧时,$BF^{2}=AF^{2}+CF^{2}+\sqrt{3}AF· CF$.
证明:当点$F$在直线$AC$左侧时,如图③,连接$AC$.

$\because BA = BC$,$∠ ABC = 60^{ \circ }$,$\therefore △ ABC$是等边三角形,
$\therefore ∠ ACB = 60^{ \circ }$,$CA = CB$,将线段$CF$绕点$C$顺时针旋转$60^{ \circ }$得到线段$CE$,连接$EF$,$EA$,则$CE = CF$,$∠ FCE = 60^{ \circ }$,
$\therefore △ CEF$是等边三角形,$\therefore ∠ CFE = 60^{ \circ }$,$FE = FC$,
$\therefore ∠ BCF = ∠ ACE$.
$\because BA = AC$,$\therefore △ BCF≌△ ACE$,
$\therefore FB = AE$.
$\because ∠ AFC = 150^{ \circ }$,$∠ CFE = 60^{ \circ }$,
$\therefore ∠ AFE = 90^{ \circ }$.
在$\mathrm{Rt} △ AEF$中,$FA^{2}+FE^{2}=AE^{2}$,
$\therefore FA^{2}+FC^{2}=FB^{2}$.
当点$F$在直线$AC$右侧时,如图④,$∠ AFC = 150^{ \circ }$,作等边三角形$FCE$,连接$AC$,$AE$.
同理可证$△ BCF≌△ ACE$,
$\therefore BF = AE$.
$\because ∠ CFE = 60^{ \circ }$,$\therefore ∠ AFE = 360^{ \circ }-150^{ \circ }-60^{ \circ }=150^{ \circ }$.

作$EM ⊥ AF$于$M$,如图⑤,在$\mathrm{Rt} △ EFM$中,
$\because ∠ EFM = 30^{ \circ }$,$\therefore EM = \dfrac{1}{2}EF$,易得$FM = \dfrac{\sqrt{3}}{2}EF$,在$\mathrm{Rt} △ AME$中,$AE^{2}=AM^{2}+EM^{2}$,
$\therefore AE^{2}=(AF+\dfrac{\sqrt{3}}{2}EF)^{2}+(\dfrac{1}{2}EF)^{2}=AF^{2}+EF^{2}+\sqrt{3}AF· EF$,
$\because AE = BF$,$EF = CF$,
$\therefore BF^{2}=AF^{2}+CF^{2}+\sqrt{3}AF· CF$.