I assume Hartshorne is talking about a Euclidean triangle, so assuming the formula for the area holds, you then want to show that $\sqrt[4]{3}$ is Hilbert constructible.
However, I think the issue here is that the field of Hilbert constructible numbers, $\Omega$, is a proper subset of $K$, the field of constructible numbers, as the square root operation is limited to $a\mapsto\sqrt{1+a^2}$, not $a\mapsto\sqrt{a}$.
Of course $\sqrt[4]{3}\in K$, since it is obtainable from rationals with the $\sqrt{}$ operation. However, take a look at Exercise 28.9 before this, which states a number $\alpha$ is constructible with Hilbert's tools if and only if $\alpha$ is constructible and totally real. Already, we know $\sqrt[4]{3}$ is constructible, but it is not totally real. We see this since the minimal polynomial of $\sqrt[4]{3}$ is $X^4-3$ which has roots $\pm\sqrt[4]{3}$ and $\pm i\sqrt[4]{3}$, so not all the conjugate elements of $\sqrt[4]{3}$ are real. Hence $\sqrt[4]{3}$ is not totally real, and thus not constructible by Hilbert's tools, although it is constructible with ruler and compass.
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 $\triangle ABC$.
Best Answer
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.