Question

如图,角a=60度,角bdf=60度,角aef=120

Answer 1

证明：取BC的中点F,设,CE,BD交于点0

因BD⊥AC,CE⊥AB,∠A = 60°

所以,EF = BC/2,DF = BC/2 ,∠BOC = ∠EOD =120°.

E B C F 四点共圆.

所以,EF = DF .∠EDB = ∠ECB,∠BDF = ∠DBF.

所以,∠EDB +∠BDF = ∠ECB+∠DBF

即：∠EDF = 180°- ∠BOC = 60°

所以,△EFD是等边三角形

所以,ED = EF = BC/2