Construction Method for a vertex-standing square inscribed within an equilateral triangle.

Question

As per the linked diagram, I am seeking a construction method for a square standing on its vertex (45 degrees orientation) that is inscribed within an equilateral triangle.

I have done my Googles without success. My straightedge and compass efforts have been entirely inadequate so I have nothing to proudly present here in that regard. I can produce it with GeoGebra (with a lot of fiddling), but it is a far throw from a classroom teachable method.

Answer 1

1. Bisect the base of the triangle.
2. Draw a perpendicular vertical at that point.
3. Bisect each of the 90° angles formed by the base and vertical.
4. Extend the bisecting rays which you drew in Step 3, until they intersect the other two legs of the triangle.
5. Set your compass at each of the intersection points created in Step 4, and set it to span the length of the ray segments (so from the intersection points from 4 to the bisecting point in Step 1). Draw circles centered at each of these points.
6. Draw segments connecting the centers of the previously drawn circles to the intersection points of the circles.

Answer 2

You can draw $60^\circ$, $90^\circ$ angles, bisect a line segment and an angle with a straightedge and compass and that's all you need.

Take a segment of any length, construct an equilateral triangle by constructing $60^\circ$ angles on it's both ends to make the triangle you have in the figure.

By symmetry, the square touches the base (in your figure) at the midpoint of the side of the triangle, so bisect the side to get the midpoint, i.e. the point where the square touches the triangle at the base.

Now, again, by symmetry, call the base of the triangle $BC$ (left side $B$), the whole triangle $\triangle ABC$, the midpoint of $BC$ as $D$, and see that the square (whose each angle is $90^\circ$) is symmetric with respect to the height $AD$ of the equilateral triangle, so it's two sides make $\dfrac{\angle BDC-90^\circ}2=45^\circ$ angles with the base $BC$, so if you construct a $45^\circ$ angle at $D$ on $BC$, you get one side of the square.