
Draw Diagonal DB. Then HE is a midline of ADB, so HE:DB = 1:2, and HEDB.
Also, GF is a midline of CBD, so GF:DB=1:2, and GFDB. Through transitivity,
GFHE, 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.
