As shown in the figure, in ladder ABCD, ad is parallel to BC, ad is less than BC, ab = DC, AC, BD intersect at point O, the angle BOC = 60 degrees, e, F, G are the midpoint of Ao, Bo, DC respectively, prove that the triangle EFG is an equilateral triangle

As shown in the figure, in ladder ABCD, ad is parallel to BC, ad is less than BC, ab = DC, AC, BD intersect at point O, the angle BOC = 60 degrees, e, F, G are the midpoint of Ao, Bo, DC respectively, prove that the triangle EFG is an equilateral triangle

Connect ed, easy to know OA = OD
∠AOD=∠BOD=60°
OE=1/2*OA=1/2*OD
According to the cosine theorem, ed = root 3 * OE, △ OED is a right triangle with ∠ OED as the right angle
ED⊥AC
G is the midpoint of the hypotenuse CD of the right triangle EDC, so CG = DG = eg
Connecting CF, we can get CF ⊥ BD, CG = DG = FG by similar method
So eg = FG = EF = 1 / 2 * AB = 1 / 2 * CD
So delta EFG is an equilateral triangle