I'll just outline the proof. You can write it out in
two-column form. The proof is mainly based on this theorem:
The segment joining the midpoints of two sides of a triangle
is parallel to the third side.
Using that theorem,
DE is parallel to BC because D and E are the midpoints of AB and AC
in triangle ABC
FH is parallel to BC because F and H are the midpoints of GB and GC
in triangle ABC
Therefore DE is parallel to FH because they are both parallel to BC
Next draw in line segment AG
DH is parallel to AG because D and H are the midpoints of AC and GC
in triangle ACG
EF is parallel to AG because E and F are the midpoints of AB and BG
in triangle ABG
Therefore DH is parallel to EF because they are both parallel to AG
Therefore DEFH is a parallelogram because both pairs of opposite
sides are parallel.
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Since the diagonals of a parallelogram bisect each other,
DG = GF, and we are given that GF = BF since F is the midpoint of BG.
Similarly, GE = GH, and we are given that GH = HC since H is the midpoint of CG.
Edwin
