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?
 A: 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.
A: HINT
► Let $C=(0,0),B=(7,0),A=(x,y)$ so the coordinates $(x,y)=\left(\dfrac{20}{7},\dfrac{8\sqrt6}{7}\right)$ (because $x^2+y^2=16$ and $(x-7)^2+y^2=25$).
► Formulas for the medians in function of sides give $(AA')^2=\dfrac14(2x16+2x25-49)=\dfrac{33}{4}$ so $AA'=\dfrac{\sqrt{33}}{2}$. Similarly $BB'=\sqrt{33}$ and $CC'=\dfrac{\sqrt{105}}{2}$.
►Coordinates of $H=(h_1,h_2)$ are given by the system $\dfrac{20-7h_2}{8\sqrt6-7h_1}$ and $AH=\dfrac{AA'}{3}$. Similarly for $I$ and $J$.
►$HI\cap{ED}$ and $HI\cap{FD}$   give  vertices $P_1$ and $P_2$ of the hexagon respectively.
$\space \space HJ\cap{FD}$ and $HJ\cap{FE}$ give vertices $P_3$ and $P_4$.
$\space \space JI\cap{FE}$ and $JI\cap{ED}$ give vertices $P_5$ and $P_6$.
►Now apply formulas for area of triangles (four times) using coordinates of points $P_i; i=1,2,3,4,5,6$
A: 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 
$$S_{\Delta EFD}-3S_{\Delta MEP}=\frac{1}{4}S_{\Delta ABC}-3\cdot\frac{1}{36}S_{\Delta ABC}=\frac{1}{6}\sqrt{8\cdot1\cdot4\cdot3}=\sqrt{\frac{8}{3}}.$$
