如图,已知等边三角形ABC中,D.E分别是BC.AC上的点,且BD=CE,连接AD.BE交于点P

因为,BD = CE ,∠ABD = ∠BCE ,AB = BC ,
所以,△ABD ≌ △BCE ,可得：∠BAD = ∠CBE ,
∠APE = ∠ABE+∠BAD = ∠ABE+∠CBE = ∠ABC = 60° .