
Triangle $ABC$ has equilateral triangles drawn on its edges.
Points $P$, $Q$ and $R$ are the centres of the equilateral
triangles. Experimentation with the interactive diagram leads to
the conjecture that $PQR$ is an equilateral triangle.
There are many ways to prove this
result. Here we have chosen two methods, one which uses only the
cosine rule and one which uses complex numbers to represent
vectors, and multiplication by complex numbers to rotate the
vectors by 60 degrees.

Another proof using a tessellation of the
plane is discussed on the Cuttheknot
website.
First the proof using the Cosine
Rule.
The sides of triangle $ABC$ are written as $a, b$ and $c$.
Centroids of equilateral triangles are at the intersection of the
altitudes so $\angle PAB$ and $\angle RAC$ are both 30 degrees.
Hence
$$AP = {2\over 3}.{\sqrt 3 c\over 2}= {c\over \sqrt 3}$$ and $$AR =
{2\over 3}.{\sqrt 3 b\over 2}= {b\over \sqrt 3}.$$
It follows that $\angle PAR = (\angle A + 60)$ degrees. By the
cosine rule
$$PR^2 = AP^2 + AR^2  2AP.AR \cos (\angle A+60) = {1\over 3}(c^2 +
b^2  2bc \cos (\angle A + 60) \quad (1).$$
Now $ \cos (\angle A + 60) = {1\over 2}\cos A  {\sqrt 3\over
2}\sin A$ and, from $\triangle ABC$: $\cos A = {b^2 + c^2  a^2
\over 2bc}$ and $\sin A = {2{\rm Area}\triangle ABC\over bc}$.
Substituting for $\cos (\angle A + 60)$ in (1) and simplifying the
expression gives:
$$ PR^2 = {1\over 3}\left[{a^2 + b^2 + c^2\over 2} + 2\sqrt 3 {\rm
Area}\triangle ABC\right].$$
This formula is completely symmetric in $a, b$ and $c$ and it
follows that $RQ^2$ and $QP^2$ have the same value and that
$\triangle PQR$ is equilateral.
Next the proof using complex numbers as
vectors.
We use $\lambda = e^{\pi i/3}$ so that $ \lambda ^2 = \lambda 
1$.
Also multiplying a complex number by $\lambda$ rotates it by 60
degrees.
Referring to the given diagram let $A, B, C$ be represented by the
complex numbers $a, b, c$. The third vertex of the equilateral
triangle drawn on $AB$ is represented by the complex number $a+
\lambda (ba)$. Therefore the centre of this triangle P is
represented by $p$ where
$$p = {1\over 3}([2  \lambda ]a +[1 +\lambda ]b).$$ Similarly $$q
= {1\over 3}(2  \lambda ]b +[1 + \lambda ]c),$$
and
$$r = {1\over 3}(2  \lambda ]c +[1 + \lambda ]a).$$
To show that $PQR$ is equilateral it is sufficient to show that $r
 q = \lambda [p  q]$ and this follows using simple algebra and
$\lambda ^ 2 = \lambda  1$.