The last paragraph at the Mathworld piece on equilateral triangles gives the answer, and cites Madachy, J. S. Madachy's Mathematical Recreations. New York: Dover, pp. 115 and 129-131, 1979.
EDIT (in response to request from Taha Akbari for more detail): Let the square have horizontal and vertical sides. Consider an equilateral triangle with one vertex at the lower left corner, $A$, of the square, and one vertex at the upper left corner, $B$, of the square, and the third vertex, $Z$, inside the square. Now consider moving the triangle vertex at $B$ to the right, toward the upper right corner, $C$, of the square, while moving $Z$ so as to keep the triangle equilateral. This increases the area of the triangle, since it increases the length of the side of the triangle, since the second vertex, $X$, of the triangle is moving away from the first vertex of the triangle.
Eventually, the triangle vertex $Z$ lies on the right side of the square, and you can't move $X$ any farther right without pushing $Z$ outside the square, so you've made the triangle as large as possible. Now the question is, why are the angles $BAX$ and $ZAD$ 15 degrees (where $D$ is the lower right corner of the square)?
The triangles $BAX$ and $ZAD$ are congruent, since $BA=AD$, $AX=AZ$, and the angles at $B$ and $D$ are equal. So the angles $BAX$ and $ZAD$ are equal. But they, together with the 60 degree angle $XAZ$, add up to the 90 degree angle $BAD$. So, they measure 15 degrees.
The area of the quadrilateral will be $$\Delta BCE-\Delta BDF$$ Now, you know $BD=2,\angle BDF=90^{\circ}$ and you can calculate $AD$ to be $3$ so that $DF=1.5$. Then you can calculate the area of $\Delta BDF$. Now, you can calculate $\angle DCE$ and $\angle EBC$ to apply the $\sin$ theorem to get $CE$. Then you can get the area of $\Delta BCE$ by $\displaystyle \frac{1}{2}BC. CE \sin \angle BCE$, and you are done.
Relevant Calculations
$$\angle DCE=\angle DCA=\tan^{-1}\left(\frac{AD}{CD}\right)=\tan^{-1}\left(\frac{3}{2}\right)\approx 56.31^{\circ}\\
\angle EBC=\angle FBC=\tan^{-1}\left(\frac{DF}{BD}\right)=\tan^{-1}\left(\frac{1.5}{2}\right)\approx 36.87^{\circ}$$
Best Answer
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 $\triangle ABC$.