An equilateral triangle is inscribed in a circle of radius $r$. If $P$ is any point on the circumference, find the value of $PA^2 + PB^2 + PC^2$. 
An equilateral triangle is inscribed in a circle of radius $r$.  If $P$ is any point on the circumference find the value of $PA^2 + PB^2 + PC^2$.  

I have managed to solve this problem  using co-ordinate geometry by taking a triangle with its centroid at (0,0) and assuming a point at $(r\cos\theta , r\sin \theta)$.  
A simpler proof using geometry however eludes me. Is it possible to prove this using Euclidean geometry or without a calculation intensive approach?  

The answer is   

$6r^2$


Without the loss of generality, we can assume that $P$ is between $A$ and $B$. Thus $m\angle{AOP}+m\angle{BOP}=120°$. Next, either $m\angle{AOP} \ge m\angle{BOP}$ or $m\angle{AOP} < m\angle{BOP}$. Let's assume $m\angle{AOP} \ge m\angle{BOP}$ (this we can do due to the symmetry of the following proof). Then $m\angle{COP}=120°+m\angle{BOP}$. 
Let $m\angle{BOP}/2=x$, then
$PA=2r\sin (60°-x)$, $PB=2r\sin x$, $PC=2r\sin (60°+x)$ and we have $$PA^2+PB^2+PC^2=4r^2 \bigl( \sin^2 (60°-x)+\sin^2 x+ \sin^2 (60°+x) \bigr)=4r^2 \bigl( \sin^2 60° \cos^2 x -2\sin 60° \cos 60° \sin x \cos x + \cos^2 60° \sin^2 x+\sin^2 x+ \sin^2 60° \cos^2 x +2\sin 60° \cos 60° \sin x \cos x + \cos^2 60° \sin^2 x \bigr)=4r^2 \bigl( \frac{3}{2} \cos^2 x +\sin^2 x+ \frac{1}{2} \sin^2 x \bigr)=6r^2$$
