Let $ABCD$ be a convex quadrilateral with $AD = BC$ and $∠A + ∠B = 120°$. Let $E$ be the midpoint of the side $CD$ and let $F$ and $G$ be the midpoints of the diagonals $AC$ and $BD$, respectively. Prove that $EFG$ is an equilateral triangle.
$E$ and $F$ are midpoints of $AC$ and $DC$, therefore, $EF = \frac{AD}{2}$ and similarly $GE = \frac{BC}{2}$. As $AD = BC$, we get $EF = EG$.
Now it is to be either proved some angle of triangle $EFG$ is is $60^\circ$ or FG = EF or GE. I am thinking it is the angle condition as there is the unused statement, $\angle A + \angle B = 120^\circ$, to use this, I tried constructing triangles $BCD$ and $ADB$ on $AD$ and $BC$ respectively but it doesn't seem to help unless I am missing something. Any help would be appreciated.
HINT.-There are many ways to choose $∠A + ∠B = 120°$ and $AD=BC$ so give you an arbitrary model to verify the property. In the attached figure, with a base $AB=6$ ,the angles are $90^{\circ}$ and $30^{\circ}$ and the equal sides measure $2$ which determines a quadrilateral. Taking letters instead of numbers you can do the same to get a general proof.