Centers of circumcircle define an equilateral triangle Let $ABC$ be a triangle with side $a,b,c$ and angles $\alpha, \beta,\gamma$.
It holds that $\alpha, \beta, \gamma<\frac{2\pi}{3}$.
At the side $\overline{BC}$ there is an equilateral triangle $\triangle BCA''$ at the outter side, i.e. $A''$  is the point for which the points $A$ and $A''$ are on different sides of the line $BC$ and for which the triangle $\triangle BCA''$ is equilateral.
Let $A'$ be the center of the circumcircle of $\triangle BCA''$.
Similarily we define the points $B'$ and $C'$ at the lines $\overline{AC}$ and $\overline{AB}$ respectively.
The radius of the circumcircle of $BCA''$ is $\frac{a}{\sqrt{3}}$.
It holds that $$|A'B'|^2=\frac{1}{3}a^2+\frac{1}{3}b^2-\frac{2}{3}ab\cos \left (\frac{\pi}{3}+\gamma\right )$$
I want to show the following :
a) $\cos \left (\frac{\pi}{3}+\gamma\right )=\frac{1}{2}\cos (\gamma )-\frac{\sqrt{3}}{2}\sin (\gamma)$
b) $A'B'C'$ is an equilateral triangle.
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First I tried to draw the above and I get :

Is that correct?
I haven't really understood how we get the angle of $\frac{\pi}{3}+\gamma$. Which is this angle in the graph?
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**EDIT: **
We have that $$|A'B'|^2=\frac{1}{3}a^2+\frac{1}{3}b^2-\frac{2}{3}ab\cos \left (\frac{\pi}{3}+\gamma\right )=\frac{1}{3}a^2+\frac{1}{3}b^2-\frac{2}{3}ab \left (\frac{1}{2}\cos (\gamma )-\frac{\sqrt{3}}{2}\sin (\gamma)$ 
\right )$$
So do we use the cosine rule also for the other angles of the triangle to get the desired result?
Or do we show in an other way that the triangle is equilateral?

Join $B'C$ and $A'C$
$\angle A'CB'=\angle A'CB+\angle BCA+\angle ACB'$
$\angle A'CB'=30^{\circ}+\gamma+30^{\circ}$
$\angle A'CB'=60^{\circ}+\gamma$
The fact used above is that the line joining circumcenter of an equilateral triangle  to any of the vertices bisects the angle corresponding to that vertex.
Now apply cosine rule on $A'CB'$
$|A'B'|^2=\frac{1}{3}a^2+\frac{1}{3}b^2-\frac{2}{3}ab\cos \left (\frac{\pi}{3}+\gamma\right )=\frac{1}{3}a^2+\frac{1}{3}b^2-\frac{2}{3}ab \left (\frac{1}{2}\cos (\gamma )-\frac{\sqrt{3}}{2}\sin (\gamma)
\right )=\frac{1}{3}(a^2+b^2-ab\cos\gamma)+\frac{2}{\sqrt 3}\Delta=\frac{1}{3}(a^2+b^2-\frac{b^2+a^2-c^2}{2})+\frac{2}{\sqrt 3}\Delta=\frac{1}{6}(a^2+b^2+c^2)+\frac{2}{\sqrt 3}\Delta$
Similarly
$|B'C'|^2=\frac{1}{3}b^2+\frac{1}{3}c^2-\frac{2}{3}bc\cos \left (\frac{\pi}{3}+\alpha\right )=\frac{1}{3}b^2+\frac{1}{3}c^2-\frac{2}{3}bc \left (\frac{1}{2}\cos (\alpha )-\frac{\sqrt{3}}{2}\sin (\alpha)
\right )=\frac{1}{3}(b^2+c^2-bc\cos\alpha)+\frac{2}{\sqrt 3}\Delta=\frac{1}{3}(b^2+c^2-\frac{b^2+c^2-a^2}{2})+\frac{2}{\sqrt 3}\Delta=\frac{1}{6}(a^2+b^2+c^2)+\frac{2}{\sqrt 3}\Delta$
Clearly above two side lengths are equal therefore.....
