(a)
Let \(M\) be the midpoint of \(\overline{BC}\), connect \(A\) to \(M\). \(\overline {AM}\) must pass through \(G\) since \(G\) is the centroid and \(\overline {AM}\) is a median of \(\triangle ABC\). In addition, connect \(D\) to \(M\), \(\overline {DM}\) must pass through \(H\) since \(H\) is the centroid and \(\overline {DM}\) is the median of \(\triangle BCD\).
\(\because \frac{MG}{MA}=\frac13=\frac{MH}{MD} \text{ and } \angle AMD = \angle AMD \therefore\) by SAS similarity, \(\triangle MGH \sim \triangle MAD\). Therefore\( \overline {GH} \parallel \overline {AD} \text{ and } AD=3GH.\) Using the same method to prove \(\triangle NKJ \sim \triangle NAD\), we can conclude \(\overline {KJ} \parallel \overline {AD} \text{ and } AD=3KJ. \because AD=3GH \text{ and } AD = 3KJ \therefore GH=KJ\), similarly \(\because \overline {KJ} \parallel \overline {AD} \text{ and }\overline {GH} \parallel \overline {AD} \therefore \overline {KJ} \parallel\overline {GH}\).
The same method is used to prove every pair of opposite sides in the hexagon are parallel and equal in length.
(b)
Let the parallel line of \(\overline{HI}\) through \(G\) and the parallel line of \(\overline{GH}\) through \(I\) intersect at . \(\because \overline{GH} \parallel \overline {OI} \text { and } \overline {GO} \parallel \overline{HI} \therefore GHIO\) is a parallelogram. Then, we can conclude \([GHI]=[GOI]\). \(\because \overline{GH} \parallel \overline {OI} \text{ and } \overline {KJ} \parallel\overline {GH} \therefore \overline {OI} \parallel \overline {KJ} \because GH = OI \text { and } GH = KJ \therefore OI = KJ\).
From this, we can conclude \(OIJK\) is also a parallelogram, which means \([IOK]=[IJK]\). \(\because \overline{HI} \parallel \overline {GO} \text{ and } \overline {HI} \parallel\overline {LK} \therefore \overline {GO} \parallel \overline {LK} \because HI = GO \text { and } HI=LK \therefore GO = LK\). From this, we can conclude \(GOKL\) is also a parallelogram, which means \([GOK]=[GLK]\). Therefore, \([GHIJKL]=2[GIK]\). Using the same method, we can conclude \([GHIJKL]=2[LHJ]\), which means that \([LHJ]=[GIK]\).
I hope this helped,
Gavin.
