如图,C点在BE上,ΔABC,ΔAED都是等边三角形,求∠DCE的度数.

∠CAD = ∠DAE+∠CAE = 60°+∠CAE = ∠BAC+∠CAE = ∠BAE ；因为,AC = AB ,∠CAD = ∠BAE ,AD = AE ,所以,△ACD ≌ △ABE ；可得：∠ACD = ∠ABE = 60°,所以,∠DCE = 180°-∠ACB-∠ACD = 60°。