Please help

(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. 

Continued here

https://web2.0calc.com/questions/geometry-polygons