18. 将矩形ABCD绕点A按顺时针方向旋转α(0° ＜ α ＜ 360°)，得到矩形AEFG.
(1) 如图，当点E在BD上时，连接DF. 求证：FD = CD.
(2) 连接GC、GB，当α为多少时，GC = GB？请说明理由，并画出图形.

18.(1)∵矩形AEFG是由矩形ABCD旋转得到的，∴AE = AB = CD，∠AEF = ∠ABC = ∠DAB = 90°，FE = BC = AD.∴∠AEB = ∠ABE.∴∠ABE + ∠EDA = 90°，∠AEB + ∠DEF = 180° - ∠AEF = 90°.∴∠EDA = ∠DEF.又∵DE = ED，AD = FE，∴$\triangle AED≌\triangle FDE$.∴AE = FD.∴FD = CD (2)当α = 60°或300°时，GC = GB　理由:若GC = GB，则点G必在BC的垂直平分线上，分两种情况讨论:①当点G在AD右侧时，如图①.设BC的垂直平分线GH交AD于点M，交BC于点H，连接DG.∵四边形ABCD是矩形，∴BC = AD，BC//AD，∠BCD = ∠ADC = 90°.∵GH垂直平分BC，∴∠CHM = 90°，CH = $\frac{1}{2}BC$.∴四边形CHMD为矩形.∴CH = DM，∠HMD = 90°.∴易得DM = $\frac{1}{2}AD$，GM⊥AD.∴GM垂直平分AD.∴AG = DG.由旋转的性质，得AD = AG，∴AD = AG = DG.∴$\triangle ADG$是等边三角形.∴∠DAG = 60°，即旋转角α = 60°.②当点G在AD左侧时，如图②.设BC的垂直平分线GH交BC于点H，交AD于点M，连接DG.同理，可得\triangle ADG是等边三角形.∴∠DAG = 60°.∴旋转角α = 360° - 60° = 300°.综上所述，当α = 60°或300°时，GC = GB.

19. 如图①，在正方形ABCD内作∠EAF = 45°，AE交BC于点E，AF交CD于点F，连接EF，过点A作AH⊥EF，垂足为H.
(1) 如图②，将△ADF绕点A按顺时针方向旋转90°得到△ABG，求证：△AGE≌△AFE.
(2) 如图③，连接BD，交AE于点M，交AF于点N. 请探究并猜想线段BM、MN、ND之间有什么数量关系，并说明理由.

19.(1)由旋转的性质，知AG = AF，∠DAF = ∠BAG，∠ABG = ∠ADF = 90°.∵四边形ABCD为正方形，∴∠ABC = 90°，∠BAD = 90°.∴G、B、E三点共线.∵∠EAF = 45°，∴∠BAE + ∠DAF = 45°.∴∠BAE + ∠BAG = 45°，即∠EAG = 45°.∴∠EAG = ∠EAF.又∵AE = AE，∴$\triangle AGE≌\triangle AFE$ (2)MN² = ND² + BM²　理由:如图，将\triangle ABM绕点A按逆时针方向旋转90°得到\triangle ADM'，连接NM'.∵四边形ABCD为正方形，∴易证∠ABD = ∠ADB = 45°，∠BAM + ∠EAD = 90°.由旋转的性质，知AM = AM'，∠ABM = ∠ADM' = 45°，∠BAM = ∠DAM'，BM = DM'.∴∠NDM' = 90°，∠DAM' + ∠EAD = 90°，即∠EAM' = 90°.∴在Rt\triangle NDM'中，M'N² = ND² + DM'².∵∠EAM' = 90°，∠EAF = 45°，∴∠MAN = ∠M'AN = 45°.又∵AN = AN，∴$\triangle AMN≌\triangle AM'N$.∴MN = M'N.又∵BM = DM'，∴MN² = ND² + BM².