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.
Let us divide ⊿ABC into the following color coded diagram.
For simplicity, we further let [yellow] = 1h, [pink] = 1k, [⊿ABC] = 630 and [green] = x
Then, [orange] = 2h and [red] = 2k.
1 : 2 = [⊿ABG] : [⊿ACG] = (1h + 2h + 1k) : (x + 2k)
After simplification, x = 6h.
∴ [yellow] $= \frac {1}{1 + 6} \cdot \frac {1}{1 + 2} \cdot 630 = 30$
The areas of the other parts can then be found and the results are shown in fig. 2.
In fig.3, it is not difficult to see that [light green] = [green] – [yellow] = … = 150
Also, [⊿HIJ] = [blue] = [red] – [light green] = … = 90, showing that [⊿HIJ] : [⊿ABC] = 1: 7
Best Answer
As $DF=DE\sqrt 2$ and the angle $\angle EDF=45^{\circ},$ the point $F$ is obtained from $E$ through the rotation composed with the homothety (common center $D$, angle and ratio as above).
Construct in this transformation the image of the side that should contain $E.$ Its intersection (if it exists) with the side that doesn't contain $D$ is $F.$