解析：
EF=DF,
证明：
过F作FM⊥AB于M,
过F作FN⊥AC于N,
过C作CM'⊥AB于M',
过A作AN'⊥BC于N',
不妨设∠BAC>∠BCA,
由∠B=60°及AD、CE是角平分线,易得
∠DFN
=∠DAN'
=(1/2)∠BAC-(90°-∠B)
=(1/2)∠BAC-30°,
∠EFM
=∠ECM'
=(90°-∠B)-(1/2)∠BCA
=30°-(1/2)∠BCA,
BF也是∠B的平分线,↔FM=FN,
∵∠DFN-∠EFM
=(1/2)∠BAC-30°-[30°-(1/2)∠BCA]
=(1/2)(∠BAC+∠BCA)-60°
=(1/2)*(180°-60°)-60°
=0,
∴∠EFM=∠DFN,
∴FE
=FM/cos∠EFM
=FN/cos∠DFN
=FD
即EF=DF
证毕!