由平行四边行ABCD中,AE⊥BD,CF⊥BD∴AE=CF,AD=BC∴△ADE≌△CBF∴BF=DE∴BE=DF再由平行四边行ABCD中得AB=CD,AB//BC∴∠ADB=∠CBD∵点G,H分别是AD,BC的中点∴BH=DG∴△BEH≌△DFG∴EH=FG,∠BEH=∠DFG∵∠HEF=180-∠BE...