在菱形ABCD中,AB=a,∠ABC=60度,点E,F分别在边CB,DC的延长线上,且∠EAF=60度求∠E=∠F 求C

⑴证明∶连接AC∵菱形ABCD ∠ABC=60°∴∠EBA=∠BCD=120°∴∠BCF=60°∵BA=BC ∠ABC=60°∴∠BAC=∠BCA=60°AB=AC∴∠BCA+∠BCF=120°即∠ACF=∠EBA=120°∵∠EAF=∠EAB+∠BAF=60°∠BAC=∠FAC+∠BAF=60°∴∠EAB=∠FAC又AB=AC ∠EBA=∠FCA∴△EBA≌△FCA （ASA)∴∠E=∠F⑵∵△EBA≌△FCA ∴EB=FC∴CE-CF=CE-EB=a