Find the area inside of a equilateral triangle Can anyone help me with this problem? I assume that similar triangles are used to solve the problem, but I can't find the solution.
Problem: ABC is an equilateral triangle with sides equal to 2 cm. BC is extended its own length to D, and E is the midpoint of AB. ED meets AC at F. Find the area of the quadrilateral BEFC in square centimeters in simplest radical form.  


Note that $\angle A = \pi/3$ because $\triangle ABC$ is equilateral. Hence we can compute:
$$ S_{\triangle ABC} = \frac{2\times 2 \sin(\pi/3)}{2} = \sqrt{3} $$
and
$$ S_{\triangle AEF} = \frac{1\times \overline{AF}\sin(\pi/3)}{2}. $$
So we have to calculate $\overline{AF}$, but it is a simple application of Menelao's Theorem:
$$1=\frac{\overline{BE}\cdot \overline{AF}\cdot \overline{CD}}{\overline{AE}\cdot\overline{FC}\cdot\overline{BD}} = \frac{1\cdot x\cdot 2}{1\cdot (2-x)\cdot 4}, $$
where $x=\overline{AF}$. Solvin this equation for $x$ gives $x=4/3$
Finally, the area $S$ we are looking for is the diference $S_{\triangle ABC}-S_{\triangle AEF}$. Then
$$ S = \sqrt{3}-\frac{4\sqrt{3}}{4\cdot 3} = \frac{2\sqrt{3}}{3}. $$
