It is proved that the connection DE, see ABCD is isosceles trapezoid, so AC=BD, AB=CD. There is AD as the common side, so the triangle ABD.

Triangle DCA so ∠ CAD = ∠ ADB = 60, so triangle AOD is an equilateral triangle.

E is the midpoint of AD, so DE⊥OA. And G is the midpoint of CD, then GE=? CD.

Similarly, GF=? BD。

E and f are OAHE respectively.

The midpoint of OB, so EF=? AB。

So EF=GE=GF, that is, the triangle EFG is an equilateral triangle.