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.  

 A: Notice that $\triangle ABC$ and $\triangle BDE$ are each half of $\triangle ABD$. Therefore $\triangle AEF$ and $\triangle CDF$ have the same area.
Now $\triangle BCF$ has the same area as $\triangle CDF$ because $BC=CD$, and $\triangle BEF$ has the same area as $\triangle AEF$ because $BE=AE$.
Therefore $\triangle AEF$, $\triangle BEF$, and $\triangle BCF$ all have the same area, and $BCFE$ is two-thirds of $\triangle ABC$.
A: 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
$$ S = \sqrt{3}-\frac{4\sqrt{3}}{4\cdot 3} = \frac{2\sqrt{3}}{3}. $$
A: Construction of side $ FB $ is crucial to calculate the three areas involved. 
Areas 
$$ p=q ;\, r= s $$
due to same base and altitude.
Apply area formula $ A= \frac12\, a\,b\, \sin C $  to find
$$ ( p+q+r) $$
and 
$$(q+r +s )$$
combination triangles. In the triangles included angle is same. First case has (2,2) as sides and other has (1,4). The common part is $(q+r) $ so the  uncommon parts $p,s$ areas are of equal area. In fact you can take cardboard to cut  flip it, to see that
$$ p=s $$
So 
$$ p =q;\, r=s;\,  p=s;\, \rightarrow \, p= q = r =s $$
In other words, $ EF,FB $ * trifurcate * the area of the given equilateral triangle $ABC$ so also $ FB,FC$ * trifurcate * $ FBD$
So the required area is $\frac23$ of the given equilateral triangle area
$$ \frac23 \frac{\sqrt3}{4} 2^2 = \frac23 \sqrt3. $$
A: $\triangle EBD = \triangle ABC -- \triangle EBD$ has 1/2 the height and 2x the base. 
$\triangle AEF = \triangle CDF$
The distance from AE to F is 1/2 the distance from AC to F.
$\triangle ABF = 2\times\triangle BCF\\
\triangle BFE = \triangle AFE$
$BCFE$ has 2/3 the area as $\triangle ABC.$
$\dfrac{2 \sqrt3}{3}$
