Help with Geometry

$\text{Just eyeballing it you've got a regular hexagon with side length equal to}\\ \text{one third the side of the triangle. Let that triangle side length be }s\\ A_t = \dfrac{\sqrt{3}}{4}s^2\\ A_h = \dfrac{3\sqrt{3}}{2}\left(\dfrac{s}{3}\right)^2 = \dfrac{\sqrt{3}}{6}s^2\\ \dfrac{A_h}{A_t} = \dfrac{\frac{\sqrt{3}}{6}}{\frac{\sqrt{3}}{4}}=\dfrac 2 3$

An easier way is to split the regular hexagon into six equilateral triangles. There are three more in the triangle $ABC$, so $\frac{6}{9}=\boxed{\frac{2}{3}}.$

Here's the picture.