Question

几何难题在正方形ABCD中,E.F分别为BC.AB上两点,且BE等于BF,过点B作AE的垂线交AC于点G,过点G作CF的

Answer 1

设FC,BG相交于J.连结DG

∵GH⊥CF,BG⊥AE

∴∠M + ∠BJC = 180°

∵∠BJC + ∠GJC = 180°

∴∠M = ∠GJC

∵ABE≌△FBC

∴∠BAE = ∠BCF

∴∠EAC = ∠FCA

∵在△MAG中,

∠M + ∠GAM + ∠AGM = 180°

在△JGC中,

∠GJC + ∠JCG + ∠JGC = 180°

∴∠AGM = ∠JGC

∵△CDG≌△CBG

∴∠DGC = ∠JGC

∴∠AGM = ∠DGC

∴D,G,H三点共线

∵∠ABC = 90°,BG⊥AE

∴∠BAE = ∠CBG

∵△CBG≌△CBG

∴∠CBG =∠CDG

∴∠BAE = ∠CBG

∴∠MAD = ∠MDA

∴AM = DM

∵△ADG≌△ABG

∴DG = GB

∵DM = DG + GM

∴AM = GB + GM