1、连接AC,‍∵∠BAD=120°,∴∠B=60°∴△ABC的等边△﹙有一个角是60°的等腰△是等边△﹚∴∠ACB=60°=∠ACF,AB=AC,又∵∠EAF=60°=∠BAC,∴∠BAE=∠CAF,∴△ABE≌△ACF﹙ASA﹚∴BE=CF；2、⑴由1、结论：△AB...