(1)证明:∵△ABC是等边三角形,
∴∠B=∠CAE=∠ACB=60°,AC=AB,
∵在△ABD和△CAE中
AB=AC
∠B=∠CAE
BD=AE
∴△ABD≌△CAE,
∴AD=CE.
(2)∵△ABD≌△CAE,
∴∠BAD=∠ACE,
∴∠DFC=∠FAC+∠ACE=∠FAC+∠BAD=∠CAE=60°.
(3)∵CG⊥AD,
∴∠CGF=90°,
∵∠DFC=60°,CF=4,
∴∠FCG=30°,
∴GF=
1
2 CF=2,
由勾股定理得:CG=2
3 .
(1)证明:∵△ABC是等边三角形,
∴∠B=∠CAE=∠ACB=60°,AC=AB,
∵在△ABD和△CAE中
AB=AC
∠B=∠CAE
BD=AE
∴△ABD≌△CAE,
∴AD=CE.
(2)∵△ABD≌△CAE,
∴∠BAD=∠ACE,
∴∠DFC=∠FAC+∠ACE=∠FAC+∠BAD=∠CAE=60°.
(3)∵CG⊥AD,
∴∠CGF=90°,
∵∠DFC=60°,CF=4,
∴∠FCG=30°,
∴GF=
1
2 CF=2,
由勾股定理得:CG=2
3 .