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