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.
I'll describe the idea I have for a straightedge-only construction. We are working in the projective plane for simplicity. You are allowed to
- Connect two points with a line.
- Find the intersection of two lines.
- Mark an arbitrary point lying on/not lying on some already constructed lines.
A construction's result should be independent of arbitrary points. Imagine as if they're supplied by an "evil goblin" and you want your result to be independent of his malice.
Let $f$ be any collineation of the projective plane. Then if your arbitrary points in a certain were $A_1, A_2, \dots A_n$ and the resulting point/line of the construction was $B$ then the "evil goblin" could have given you the arbitrary points $f(A_1), f(A_2), \dots f(A_n)$ and these result would have been $f(B)$.
Therefore you can only construct projective invariants of the initially given points. You cannot, then, halve an interval, as that is not a projective invariant of the two endpoints. This is because collineations act 3-transitively on a line.
Having three points with given distances on a line (one may be at infinity, this is equivalent to the ability to construct parallel lines) is enough for the projective invariance to go away. We can use this construction: https://en.wikipedia.org/wiki/Cross-ratio#/media/File:Pappusharmonic.svg to get the harmonic conjugate point on the line and from there we can construct any rational distance. With duality this extends to angles, resulting in right angles, so we pretty much regain Euclidean geometry.
This is considered folklore among Hungarian students, mostly thanks to Lajos Pósa.
Best Answer
If you have a segment $AB$, place the unit length segment on the line where $AB$ lies, starting with $A$ and in the direction opposite to $B$; let $C$ be the other point of the segment. Now draw a semicircle with diameter $BC$ and the perpendicular to $A$; this line crosses the semicircle in a point $D$. Now $AD$ is the square root of $AB$.
$\triangle BCD$ is a right triangle, like $\triangle ACD$ and $\triangle ABD$; all of these are similar, so you find out that $AC/AD = AD/AB$. But $AC=1$, so $AD = \sqrt{AB}$.
See the drawing below: