延长DC,并与BF交于G.在△ABE与△CGF中,∵AE‖＝CF,AB‖CG,∴∠BAE=∠GCF,∠AEB=∠CFG,∴△ABE≌△CGF.∴∠ABE=∠CGF.AB=CG.∴△BCG是直角等腰三角形,∴∠CGB=45°.作CH⊥BF于H,则CH=1/2BG.又FC=AC=BG,∴CH=1/2FC,又...