证明:(1) ∵AD//BC, ∴∠DAE=∠F. 又∠AED =∠ CEF,DE=EC ∴△ADE≌△FCE. ∴AE= EF,AD=FC. ∴AD+BC=CF+BC=BF. 又AD + BC =AB. ∴AB = BF. ∴BE是等腰△ABF底边上的中线. ∴BE平分∠ABC(三线合一). (2) ∵AB=BF. ∴F=BAE. 又∠F=∠DAE,∠BAE=∠DAE, ∴AE平分∠BAD. ∴∠BAE+∠ABE= (∠BAD+∠ABC). 又∵AD//BC, ∴∠BAD+∠ABC= 180°. ∴∠BAE+∠ABE=90°. ∴∠AEB=90°. ∴AE⊥BE. |