在等边三角形ABC中D、F分别为CB、BA上的点且CD=BF以AD为边作等边三角形ADE求证四边形CDEF是平行四边形

连结EB
AE = AD
AB = AC
∠EAB + ∠BAD = ∠BAD + ∠DAC = 60°
∴ ∠EAB = ∠DAC
∴△AEB ≌△ADC
∴EB = DC,∠EBF = ∠ACD = 60°
∵BF = CD
∴EB = BF
∴△EBF是等边三角形
∴ ∠EFB = ∠ABC = 60°,EF = FB
∵∠ABC = ∠ACB,FB = CD
∴∠EFB = ∠ACB
∴FB // CD
∴四边形EFCD为平行四边形