Construct an equilateral triangle with area equal to a given triangle It is straightforward to construct (straight-edge and compass) an isosceles triangle with area equal to a given triangle $\triangle ABC$, for instance as follows:


*

*Construct the line through $A$ parallel to $BC$ (demonstration of method);

*Construct the perpendicular bisector of $BC$ (demonstration of method);

*The perpendicular from (2) meets the line from (1) at $D$; draw $\triangle DBC$.


This triangle is isosceles, since $D$ is equidistant from $B$ and $C$, and has the same base and altitude as $\triangle ABC$ so has the same area.
Suppose we want to go further and construct an equilateral triangle with area equal to the given triangle — how might we go about that?

 A: A possible construction.



*

*Draw $DE$ through $A$ in such a way that $DECB$ is a rectangle;

*Take $Q$ as the intersection of the angle bisectors of $\widehat{BDE}$ and $\widehat{DEC}$;

*Take $T$ on $CE$ such that $TE=EQ$;

*Take $F$ on $DT$ such that $DF=DE$;

*Let $\ell$ be the perpendicular to $BE$ through $E$;

*Take $G$ on $\ell$ such that $BG=BF$.


$EG$ is the side of an equilateral triangle with the same area of $ABC$.
A: 
The fewer steps the better; I found I had to use the geometric mean construction and I wonder whether this step might be avoidable.

Something like a geometric mean is unavoidable since the problem is a quadratic equation for the side length of the equilateral triangle. 
Here is a relatively efficient construction using a geometric mean.  The savings is in re-using one side of ABC as the base of the equilateral triangle, and in using a non-perpendicular line to measure altitude, and allowing the semicircle construction of geometric mean to be applied.
On one of the sides, say AB, build an equilateral triangle ABD. Extend line CD to intersect AB at P. Find a length $g$ equal to the geometric mean of PC and PD, and take a point H on PD with PH = $g$.  The equilateral triangle whose "height" measured along line PCD (from P) is $g$ has the same area as the given triangle.
The construction of the geometric mean is made easier by extending line CD to E so that P is the midpoint of EC, then using ED as diameter of a circle and taking $g=|PG|$ for PG a perpendicular to CD with G on the circle.   Then draw the parallel to AB through H and intersect it with AD (at point K) to cut off the correct side length (AK) of equilateral triangle.
A: Thank you to all you brilliant mathematicians for your ingenious methods on this topic. I do not understand trigonometry very well because I am not very well educated, but this is Euclid's approach to the problem which is general for generating similar figures of any kind of a given area, which, although perhaps cumbersome, I found convenient and readily intelligible.
An equilateral triangle is made, a parallelogram constructed equal to it, another equal to the desired area applied alongside it in the parallels, then a mean proportional taken between the lengths of the two parallelograms. Because as the first line is to the third, so is the figure on the first to the similarly described figure on the second, the equilateral described on the mean proportional contains the desired area. 
The other methods are simpler I just thought I'd post this because it's purely geometrical. Below is a link to the proposition and my construction.
http://aleph0.clarku.edu/~djoyce/elements/bookVI/propVI25.html
construction
A: Using Heron's formula:
$$ \sqrt{s(s-a)(s-b)(s-c)}=\frac{a^2 \sqrt{3}}{4}  $$
$$\Rightarrow$$
$$a^2=4\sqrt{\frac{s}{3} (s-a)(s-b)(s-c)}$$
with subistitution:$$y=\sqrt{\frac{s}{3}(s-a)}, \quad z=\sqrt{(s-b)(s-c)}$$
$$\Rightarrow$$
$$a=2\sqrt{yz}$$
A: Let the required equilateral triangle have side $s$, and the given triangle have base $b$ and altitude $h$, so:
$$ \frac{1}{2}bh = \frac{1}{2}s^2 \sin 60^\circ \implies s^2 = b \left( \frac{h}{\sin 60^\circ} \right)$$
Therefore we can construct $s$ as the geometric mean of $b$ and $\frac{h}{\sin 60^\circ}$.



*

*Draw the perpendicular from $A$ to $BC$ (produced if necessary); let the foot of the perpendicular be $F$.

*Draw an arc center $A$ through $F$ and another centre $F$ through $A$; these intersect at $P$ so that $\angle FAP = 60^\circ$.

*Draw the angle bisector of $\angle FAP$; let $Q$ be its intersection with $BC$ (produced if necessary), so that $\angle FAQ = 30^\circ$.  


Considering $\triangle AFQ$ we have $\sin 60^\circ = \cos 30^\circ = \frac{AF}{\color{red}{AQ}}$, so that $\color{red}{AQ} = \frac{AF}{\sin 60^\circ}$. We need to take the geometric mean of this length and the base.



*Draw an arc, center $C$ and radius $\color{red}{AQ}$, to construct $R$ on $BC$ produced such that $\color{red}{AQ} = \color{red}{CR}$.

*Construct the midpoint $O$ of $BR$, and hence draw a circle with diameter $BR$.

*Construct the perpendicular to $BR$ through $C$; let $S$ be its intersection with the circle from step (5). 


Now by considering similar triangles $\triangle BCS$ and $\triangle SCR$, or bearing in mind that $\color{green}{CS}$ is one half of a chord bisected by the diameter $BR$ and then applying the intersecting chords theorem, we have $\color{green}{CS}^2 = BC \cdot \color{red}{CR}$. Hence $\color{green}{CS}$ is the required geometric mean, and is the side of an equilateral triangle with area equal to that of the given triangle.

A: I wondered whether it was possible to do this by "morphing" an isosceles triangle into an equilateral one — here's an approach that creates an equilateral triangle by moving the apex of the isosceles triangle, then scales its sides to restore the original area.
Without loss of generality, assume $\triangle ABC$ is isosceles with $AB=AC$. (Otherwise, construct an isosceles triangle with area equal to the given triangle, e.g. by the method given in the question.)



*

*Draw a circle center $B$ through $C$, and another center $C$ through $B$.

*Take $D$ as the intersection of these circles that lies the opposite side of $BC$ to $A$; $\triangle BCD$ is equilateral.

*Draw $AD$ and take $M$ to be its intersection with $BC$; $M$ is the midpoint of $BC$.


Although $\triangle BCD$ is equilateral, it isn't the triangle we need because it does not have the same area as $\triangle ABC$. They share base $BC$, so the ratio of their areas is simply the ratio of their altitudes:
$$ \frac{\text{Area } \triangle ABC}{\text{Area } \triangle BCD} = \frac{AM}{MD}$$
So to obtain the desired equilateral triangle we must scale the lengths of $\triangle BCD$ by $\sqrt \frac{AM}{MD}$.



*Construct $O$ as the midpoint of $AD$ and hence draw a circle diameter $AD$.

*Produce $BC$ to meet the circle from (5) at $E$, then by the intersecting chords theorem we have $EM^2 = AM \cdot MD$ and hence $\tan \angle EDM = \frac{EM}{MD} = \sqrt \frac{AM}{MD}$. 

*Take $P$ on $AD$ such that $PD = BC$.

*Construct the perpendicular to $AD$ through $P$, and take $Q$ as its intersection with $DE$.


Then $PQ$ is the side of an equilateral triangle with area equal to the given triangle, since $$\frac{PQ}{BC} = \frac{PQ}{PD} = \tan \angle EDM = \sqrt \frac{AM}{MD} =  \sqrt \frac{\text{Area } \triangle ABC}{\text{Area } \triangle BCD}$$

