Question

在四边形ABCD中,分别以AB、AD为边向外作等边三角形ABE、三角形ADF，延长CB交AE于点G，点G在点A、E之间，

连结EF、EC。请证明三角形EAF全等于三角形EBC。

Answer 1

四边形ABCD必须是平行四边形才成立

①AF=AD=BC

   AB=EB

∠EDC=360°-∠ABC-∠ABE=300°-∠ABC

∠EAF=120°+∠BAD=120°+(180°-∠ABC)=300°-∠ABC

△EBC≌△EAF

Answer 2

等边三角形ABE、三角形ADF
AF=AD=BC
AB=AE
∠FAE=∠FAD+∠DAB+∠EAB=120°+∠DAB
∠EBC=360°-∠ABC-∠ABE=360°-（180°-∠DAB）-60°=120°+∠DAB
所以三角形EAF全等于三角形EBC

Answer 3

AF=AD=BC
AB=EB
∠EDC=360°-∠ABC-∠ABE=300°-∠ABC
∠EAF=120°+∠BAD=120°+(180°-∠ABC)=300°-∠ABC
△EBC≌△EAF