Question

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

Answer 1

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=∠F

⑵∵△EBA≌△FCA

∴EB=FC

∴CE-CF=CE-EB=a