How to prove that the sum of squares of the distances from $A,B,C$ to a random tangent is constant. Given an equilateral triangle $ABC$ inscribing circle(O). How to prove that the sum of squares of the distances from $A,B,C$ to a random tangent is constant. Is there a simple way?
 A: Here's the claim ...
Proposition: Let $T$  be an equilateral triangle, let $P$ be a point on the circumcircle  of the triangle, and let $l_P$ be the line tangent to the circle at $P$. Then as $P$ varies, the sum of the distances to $l_P$  from the vertices of the triangle remains constant.
Such a simple claim should have a natural, synthetic, geometric proof, but I couldn't find one, so instead, I bashed it out algebraically.  Here are the gory details ...
Without loss of generality, assume the circumcircle is the standard unit circle in $R^2$, and the vertices of $T$, in complex form, are the cube roots of unity. 
Letting $k = (2\pi)/3$, the vertices, in coordinate form, are:
$$U = (1,0)$$
$$V = (\cos(k),\sin(k))$$
$$W = (\cos(k),-\sin(k))$$
Let $P$ be an arbitrary point on the circumcircle. Then $P = (\cos(t),\sin(t))$.
The line $l_P$  has the equation $ax + by  = 1$ where $a = \cos(t)$ and $b = \sin(t)$.
By the formula for the distance from a point to a line, 
$$d(U,l_p) = |\cos(t) - 1| = 1 - \cos(t)$$
$$d(V,l_p) = |\cos(t) \cos(k) + \sin(t) \sin(k) - 1| = |\cos(t-k) - 1| = 1 - \cos (t-k)$$
$$d(W,l_p) = |\cos(t) \cos(k) - \sin t \sin k - 1| = |\cos(t+k) - 1| = 1 - \cos(t+k)$$
hence the sum of distances is
$$3 -  (\cos(t) + \cos(t - k) + \cos(t + k))$$
which is equal to 3 since $$\cos(t) + \cos(t - k) + \cos(t + k) = (\cos(t))(1 + 2\cos(k)) = 0$$
A: 
Without loss of generality, we may assume to be in the depicted configuration.
By the parallel axis theorem the sum $PA^2+PB^2+PC^2$ does not depend on the position of $P$ on the circumcircle of $ABC$, it is just $6R^2$. On the other hand, by the Pythagorean theorem such sum equals
$$ (AA'^2+BB'^2+CC'^2)+(PA'^2+PB'^2+PC'^2) $$
or
$$ (AA'^2+BB'^2+CC'^2)+(OA'^2+OB'^2+OC'^2)-3R^2. $$
Since $O$ is the centroid of $ABC$, $P$ is the centroid of $A',B',C'$. Additionally, Ptolemy's theorem ensures that $PB=PA+PC$. Can you finish from here?
