Area of the intersection of two triangles. Let $\triangle{ABC}$ be a triangle with $AB=5$, $BC=7$, and $CA=4$. Define $D$, $E$, and $F$, to be the midpoints of $AB$, $BC$, and $CA$ respectively. Let $G$ the intersection of the medians of $\triangle{ABC}$, and let $H$, $I$, and $J$ be the midpoints of $AG$, $BG$, and $CG$ respectively. Find the area of the hexagonal region common to both $\triangle{DEF}$ and $\triangle{HIJ}$. 

This is too tough for me to even start. I suppose you would start with similar triangles? I know for a fact that the areas of both $\triangle{DEF}$ and $\triangle{HIJ}$ are both a quarter the area of $\triangle{ABC}$, but I have no idea how to proceed. Can somebody help?
We can start by finding the length of the median, using Apollonius' Theorem, which yields $AB^2 + AC^2 = 2(AE^2 + BD^2) \rightarrow 25+16 = 2(AE^2 + 12.25) \rightarrow AE = \frac{\sqrt{33}}{2}$. It is easy to see that the ratio of $AG : GE = 2:1$ , and since $H$ is the midpoint of $AG$, we know that $AE$ is trisected by $H, G$. Repeat this for the other medians, and then it should be easy to go from there.
The hint.

Let $AE\cap IJ=\{T\}$, $IJ\cap ED=\{M\}$ and $EF\cap IJ=\{P\}$.

Thus, $T$ a midpoint of $EG$, which says $$ET=\frac{1}{2}\cdot\frac{1}{3}AE=\frac{1}{6}AE.$$
Also, since $ADEF$ is a parallelogram and $IJ||DF,$ we obtain $ET$ is a median of $\Delta MEP.$

But $\Delta MEP\sim\Delta CAB,$ which gives
$$S_{\Delta MEP}=\frac{1}{36}S_{\Delta ABC}.$$
Id est, the needed area is equal to 
$$S_{\Delta EFD}-3S_{\Delta MEP}=\frac{1}{4}S_{\Delta ABC}-3\cdot\frac{1}{36}S_{\Delta ABC}=\frac{1}{6}\sqrt{8\cdot1\cdot4\cdot3}=\sqrt{\frac{8}{3}}.$$