答案：
6. C 解析：$\because$四边形$ABCD$是正方形，$\therefore AC⊥ BD$，$AB = BC = AD$，$OA = OB = OC = OD$，$∠ ABC = ∠ ABF + ∠ CBG = 90^{\circ}$，$∠ ABO = ∠ BCO = 45^{\circ}$，$∠ AOB = 90^{\circ}$。$\because BF⊥ AE$，$CG⊥ BF$，$\therefore ∠ AFB = ∠ BGC = 90^{\circ}$，$∠ CBG + ∠ BCG = 90^{\circ}$，$\therefore ∠ ABF = ∠ BCG$。在$△ ABF$和$△ BCG$中，$\begin{cases}∠ ABF = ∠ BCG,\\∠ AFB = ∠ BGC,\\AB = BC,\end{cases}$ $\therefore △ ABF≌△ BCG(\mathrm{AAS})$，$\therefore AF = BG$，①正确；$\because$正方形的对角线平分每组对角，$\therefore ∠ BAC = ∠ OBC = 45^{\circ}$。由$△ ABF≌△ BCG$得$∠ BAF = ∠ CBG$，$\therefore ∠ FAO = ∠ GBO$。又$\because OA = OB$，$\therefore △ FAO≌△ GBO$，$\therefore FO = GO$，$∠ FOA = ∠ GOB$。$\because ∠ AOG + ∠ GOB = 90^{\circ}$，$\therefore ∠ AOG + ∠ FOA = 90^{\circ} = ∠ FOG$，故$△ GOF$为等腰直角三角形，故②正确；$\therefore ∠ FGO = 45^{\circ}$。$\because ∠ BGC = 90^{\circ}$，$\therefore ∠ OGC = 45^{\circ} = ∠ FGO$，$\therefore GO$平分$∠ FGC$，故⑤正确；设$GC$交$BD$于点$I$，连接$HI$，如图①所示，$\because △ GOF$为等腰直角三角形，$∠ OGC = 45^{\circ} = ∠ FGO$，$\therefore ∠ HFO = ∠ IGO = 45^{\circ}$。在$△ FOH$和$△ GOI$中，$\begin{cases}∠ HFO = ∠ IGO,\\FO = GO,\\∠ FOH = ∠ GOI,\end{cases}$ $\therefore △ FOH≌△ GOI(\mathrm{ASA})$，$\therefore HO = IO$，$GI = HF$，$\therefore △ HOI$为等腰直角三角形，$\therefore HI^{2} = 2HO^{2}$，在$Rt△ HGI$中，由勾股定理可得$HG^{2} + GI^{2} = HI^{2}$，即$HG^{2} + HF^{2} = HI^{2} = 2HO^{2}$，故④正确；
如图②所示，作$GJ = GI$，$\because GI = HF$，$\therefore GJ = HF$，$\therefore BG + HF = BG + GJ = BJ$，当且仅当$BG = HJ$时，$BG + HF = GH$成立，故③不一定正确。
综上，正确的序号为①②④⑤，故选C。

答案：
7. C 解析：如图①，连接$DC$。$\because AC = BC$，$∠ ACB = 90^{\circ}$，点$D$为$AB$边的中点，$\therefore ∠ B = 45^{\circ}$，$∠ DCE = \frac{1}{2}∠ ACB = 45^{\circ}$，$CD⊥ AB$，$CD = \frac{1}{2}AB = BD$，$\therefore ∠ DCE = ∠ B$，$∠ CDB = 90^{\circ}$。$\because ∠ EDF = 90^{\circ}$，$\therefore ∠ CDE = ∠ BDF$。在$△ CDE$和$△ BDF$中，$\begin{cases}∠ CDE = ∠ BDF,\\CD = BD,\\∠ DCE = ∠ B,\end{cases}$ $\therefore △ CDE≌△ BDF(\mathrm{ASA})$，$\therefore CE = BF$，故①正确；$\because △ CDE≌△ BDF$，$\therefore ∠ BFD = ∠ DEC$，$DF = DE$，$\therefore ∠ BFD + ∠ DFC = 180^{\circ} = ∠ DEC + ∠ DFC≠∠ DEC + ∠ DBF$，故②错误；$\because ∠ EDF = 90^{\circ}$，由$△ CDE≌△ BDF$知$DE = DF$，$\therefore EF^{2} = DE^{2} + DF^{2} = 2DE^{2}$，故③正确；如图②，连接$CD$，同理可证$△ DEC≌△ DFB$，$∠ DCE = ∠ DBF = 135^{\circ}$。$\because S_{△ DEF} = S_{△ CFE} + S_{△ DBC} = S_{△ CFE} + \frac{1}{2}S_{△ ABC}$，$\therefore S_{△ DEF} - S_{△ CEF} = \frac{1}{2}S_{△ ABC}$，故④正确。故选C。

答案：
8. (1)不能 解析：由题意得$BE = 4t\ \mathrm{cm}$，$BF = 3t\ \mathrm{cm}$，$\because BE≠ BF$，$\therefore$四边形$EBFB'$不能是正方形。
(2)如图①，连接$EG$，$\because$四边形$ABCD$是矩形，$\therefore BE// CG$，$∠ ABC = ∠ C = 90^{\circ}$。$\because BE = CG = 4t\ \mathrm{cm}$，$\therefore$四边形$BCGE$是平行四边形。$\because ∠ C = 90^{\circ}$，$\therefore$平行四边形$BCGE$是矩形，$\therefore EG = BC$，$EG// BC$。$\because M$，$N$分别是$EF$，$FG$的中点，$\therefore MN = \frac{1}{2}EG = \frac{1}{2}BC$，$EG// MN// BC$，$\therefore MN// BF$，当$MN = BF$时，四边形$BMNF$是平行四边形，此时$BF = \frac{1}{2}BC$，即$3t = \frac{1}{2}×8$，解得$t = \frac{4}{3}$，故当$t = \frac{4}{3}$时，四边形$BMNF$是平行四边形。

(3)存在实数$t$，使得点$B'$与点$O$重合，如图②，连接$B'B$交$EF$于点$H$，连接$AC$，$BD$，$\because$四边形$ABCD$为矩形，$AB = 6\ \mathrm{cm}$，$BC = 8\ \mathrm{cm}$，$\therefore AC = BD = \sqrt{6^{2} + 8^{2}} = 10(\mathrm{cm})$，$\therefore BO = \frac{1}{2}BD = 5\ \mathrm{cm}$。$\because △ EBF$关于直线$EF$的对称图形是$△ EB'F$，$\therefore EF$是线段$B'B$的垂直平分线，$\therefore BH = B'H$，当点$B'$与点$O$重合时，$BH = B'H = \frac{1}{2}×5 = \frac{5}{2}(\mathrm{cm})$。在$Rt△ BEF$中，$BH⊥ EF$，$BE = 4t\ \mathrm{cm}$，$BF = 3t\ \mathrm{cm}$，$\therefore EF = \sqrt{BE^{2} + BF^{2}} = 5t\ \mathrm{cm}$。$\because BH = \frac{5}{2}\ \mathrm{cm}$，$\therefore S_{△ BEF} = \frac{1}{2}EF· BH = \frac{1}{2}BE· BF$，即$\frac{1}{2}×5t×\frac{5}{2} = \frac{1}{2}×4t×3t$，解得$t = \frac{25}{24}$。

答案：
9. (1)② 解析：当邻余四边形是同一底边上是$45^{\circ}$的等腰梯形时，它是轴对称图形，因为邻余四边形的邻角互余，所以它不能是平行四边形，所以它不能是中心对称图形，故答案为②。
(2)如图①，作$DE// BC$交$AB$于$E$，$\therefore ∠ AED = ∠ B$。$\because AB$是邻余线，$\therefore ∠ A + ∠ B = 90^{\circ}$，$\therefore ∠ A + ∠ AED = 90^{\circ}$，$\therefore ∠ ADE = 90^{\circ}$。$\because AB// CD$，$\therefore$四边形$CDEB$是平行四边形，$\therefore DE = BC$，$BE = CD = 2$。$\because AB = 7$，$\therefore AE = AB - BE = 7 - 2 = 5$。$\because AD = 3$，$\therefore DE = \sqrt{AE^{2} - AD^{2}} = \sqrt{5^{2} - 3^{2}} = 4$，$\therefore BC = 4$。

(3)①如图②，作射线$AN$，并截取$NG = AN$，连接$CG$，$BG$，$MN$，设$AD$和$BC$的延长线交于$H$，$\because$点$N$是$CD$的中点，$\therefore DN = CN$。$\because ∠ AND = ∠ CNG$，$AN = NG$，$\therefore △ ADN≌△ GCN(\mathrm{SAS})$，$\therefore CG = AD = a$，$∠ DAN = ∠ CGN$，$\therefore AD// CG$，$\therefore ∠ HCG = ∠ AHC$。$\because M$是$AB$的中点，$NG = AN$，$\therefore MN = \frac{1}{2}BG$。$\because AB$是邻余线，$\therefore ∠ DAB + ∠ CBA = 90^{\circ}$，$\therefore ∠ AHC = 90^{\circ}$，$\therefore ∠ HCG = 90^{\circ}$，$\therefore ∠ BCG = 90^{\circ}$。$\because AD = a = CG$，$BC = b$，$\therefore BG = \sqrt{CG^{2} + BC^{2}} = \sqrt{a^{2} + b^{2}}$，$\therefore MN = \frac{\sqrt{a^{2} + b^{2}}}{2}$，$\therefore S_{\mathrm{正方形}EMFN} = \frac{1}{2}EF· MN = \frac{1}{2}MN^{2} = \frac{a^{2} + b^{2}}{8}$。
②$\sqrt{a^{2} + b^{2}} ＜ AB≤\sqrt{a^{2} + b^{2}} + c$ 解析：如图②，连接$HN$，$HM$，由①知$∠ AHC = 90^{\circ}$，$MN = \frac{\sqrt{a^{2} + b^{2}}}{2}$，$\because$点$N$是$CD$的中点，$M$是$AB$的中点，$\therefore HN = \frac{1}{2}CD = \frac{1}{2}c$，$HM = \frac{1}{2}AB$。$\because$邻余四边形$ABCD$中，$AB$是邻余线，$\therefore$点$H$在$CD$上方，$\therefore MN ＜ HM≤ MN + HN$，$\therefore \frac{\sqrt{a^{2} + b^{2}}}{2} ＜ HM≤\frac{\sqrt{a^{2} + b^{2}}}{2} + \frac{1}{2}c$，$\therefore \sqrt{a^{2} + b^{2}} ＜ AB≤\sqrt{a^{2} + b^{2}} + c$。