(1)证明:∵△ABC是等边三角形,
∴∠A=∠B=∠C=60°,AC=AB=BC,
∵AD=BE=CF,
∴AC-CF=BC-BE=AB-AD,
∴EC=AF=BD,
∴在△ADF,△BED,△CFE中,
,
∴△ADF≌△BED≌△CFE(SAS),
∴DF=DE=EF,
∴△DEF是等边三角形,
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/bd3eb13533fa828b70b77f52fe1f4134960a5aef?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
(2)解:(1)的逆命题成立,
已知:△DEF是等边三角形,求证:AD=BE=CF.
证明:∵△DEF是等边三角形,
∴∠EDF=∠EFD=∠DEF=60°,DF=EF=DE,
∵等边三角形ABC,
∴∠A=∠B=∠C=60°,
∴∠ADF+∠AFD=120°,∠ADF+∠BDE=120°,∠BDE+∠DEB=120°,∠AFD+∠EFC=120°,
∴∠ADF=∠DEB=∠EFC,
在△ADF,△BED,△CFE中,
∵
| DF=ED=FE | ∠A=∠B=∠C | ∠ADF=∠BED=∠CFE |
| |
,
∴△ADF≌△BED≌△CFE(AAS),
∴AD=BE=CF.