Analysis: Let $ABC$ be the given triangle and let $2x$ be the length of the sides of the required equilateral triangle $XYZ$. Let $AD$ be an altitude of $ABC$, $h = |AD|$ and $a = |BC|$. The area of $ABC$ is equal to $ah/2$ and the area of $XYZ$ is $\sqrt{3}x^2$. Then we must have $ah/2 = \sqrt{3}x^2$, i.e.
$$x^2 = \frac{a}{2} \cdot \frac{h}{\sqrt{3}}.$$
The natural idea then is to construct the length $h/\sqrt{3}$ as half the side length of an equilateral triangle with height $h$.

Construction: Draw the altitude through $A$ with foot $D$. Through $A$ draw a line parallel to $BC$ and draw a line making an angle of $60^\circ$ with $BC$ at $D$ and let these two lines intersect at $E$. Since $\angle DAE = 90^\circ$ and $\angle ADE = 30^\circ$, $AE = h/\sqrt{3}$. Extend $AE$ to $F$ so that $|EF| = |BC|/2 = a/2$. Now construct a circle on $AF$ as diameter and erect a perpendicular at $E$ meeting the circle at $Y$ and $Z$. Finally construct an equilateral triangle $XYZ$ on $YZ$. This is the required triangle.

Justification: From the power of the point $E$ with respect to the circle we see $|YE|^2 = |YE| \cdot |EZ| = |AE| \cdot |EF| = h/\sqrt{3} \cdot a/2 = x^2$ and so $|YE| = x$ and $|YZ| = 2x$ which shows that $XYZ$ is the required triangle.