Question 207598
{{{drawing(400,254.5454545,-1,10,-1,6,
triangle(0,0,4.1934,2.7232,3.1480,0),
triangle(3.1480,0,4.1934,2.7232,4.8942,1.8155),
triangle(3.1480,0,5.5951,.90773,4.8942,1.8155),
triangle(3.1480,0,6.2960,0,4.8942,1.8155),
triangle(6.6405,3.6309,4.1934,2.7232,4.8942,1.8155),
triangle(6.6405,3.6309,5.5951,.90773,4.8942,1.8155),
triangle(8.3867,5.4464,4.1934,2.7232,4.8942,1.8155),
triangle(4.1934,2.7232,8.3867,5.4464,6.2960,0),
locate(0,0,A),locate(6.2960,0,B),locate(8.3867,5.4464,C),
locate(4.1934,2.7232-.2,D),locate(3.1480,0,E),locate(5.5951-.2,.90773-.2,F),
locate(4.8942-.2,1.8155-.2,G),locate(6.6405,3.6309,H)
  )}}}
<pre><font size = 4 color = "indigo"><b>
I'll just outline the proof.  You can write it out in
two-column form.  The proof is mainly based on this theorem:

<i>The segment joining the midpoints of two sides of a triangle
is parallel to the third side.</i>

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

{{{drawing(400,254.5454545,-1,10,-1,6,
triangle(0,0,4.1934,2.7232,3.1480,0),
triangle(3.1480,0,4.1934,2.7232,4.8942,1.8155),
triangle(3.1480,0,5.5951,.90773,4.8942,1.8155),
triangle(3.1480,0,6.2960,0,4.8942,1.8155),
triangle(6.6405,3.6309,4.1934,2.7232,4.8942,1.8155),
triangle(6.6405,3.6309,5.5951,.90773,4.8942,1.8155),
triangle(8.3867,5.4464,4.1934,2.7232,4.8942,1.8155),
triangle(4.1934,2.7232,8.3867,5.4464,6.2960,0),
locate(0,0,A),locate(6.2960,0,B),locate(8.3867,5.4464,C),
locate(4.1934,2.7232-.2,D),locate(3.1480,0,E),locate(5.5951-.2,.90773-.2,F),
locate(4.8942-.2,1.8155-.2,G),locate(6.6405,3.6309,H),
line(0,0,4.8942,1.8155) )}}}
 
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.

-----------------------

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</pre>