Draw Diagonal DB. Then HE is a midline of ADB, so HE:DB = 1:2, and HE||DB.
Also, GF is a midline of CBD, so GF:DB=1:2, and GF||DB. Through transitivity,
GF||HE, and GF=HE. Applying the same argument to the opposite pair of sides,
we can see that EFGH is a parallelogram. A representation of one of the triangles
and its midline can be seen on the left.