Ratio of Triangular Areas

By the symmetry of the given figure, $HI\parallel BC \parallel DE$ . We know that for an equilateral triangle, the median drawn is your perpendicular bisector as well as your angle bisector.

Therefore, $\angle ECG$ = $60^{o}$ = $\angle IEG$

Let each side of $\triangle ABC$ = a units

Thus, $IE$ = $GE \cos 60^{o}$ = ( $EC \sin 60^{o}$ ) x $\cos 60^{o}$ = $\frac{a}{2}$ x $\frac{\sqrt3}{2}$ x $\frac{1}{2}$ = $\frac{\sqrt3 a}{8}$

$J$ is the centroid that divides $BE$ in the ratio $2:1$ . Thus, $JE$ = $\frac{1}{3}$ x $\frac{\sqrt3 a}{2}$ = $\frac{a}{2 \sqrt3}$

$JI$ = $JE$ - $IE$ = $\frac{a}{8 \sqrt3}$

$\triangle HIJ \sim \triangle CBJ$ by AA similarity.

$\dfrac{Ar\triangle HIJ}{Ar\triangle CBJ}$ = $\frac{JI^{2}}{BJ^{2}}$ = $\frac{a^{2}}{(192 a^{2})/3}$ = $\frac{1}{64}$

We know that by symmetry, $Ar\triangle CBJ$ = $\frac{1}{3}$ x $Ar\triangle ABC$

Hence, $\dfrac{Ar\triangle HIJ}{Ar\triangle ABC}$ = $\frac{1}{3}$ x $\frac{1}{64}$ = $\frac{1}{192}$