Triangle Sides/orthocenter/vertices problem I have to prove that:
The sum of the squares of the distances of the vertices
of a triangle from the orthocenter is equal to twelve times the square of the
circumradius diminished by the sum of the squares of the sides of the
triangle. Specifically, denote $A,B,C,H$ be the vertices of a triangle, and the orthocenter respectively. And let $R$ be the circumradius of $\triangle ABC$, prove:
$$AH^2+BH^2+CH^2 = 12R^2 - AB^2 - BC^2 - AC^2$$
Any hints/tips? Kinda lost here.
 A: By Euler's theorem, in any triangle $ABC$ the circumcenter $O$, the orthocenter $H$ and the centroid $G$ are collinear and $HO=3\cdot GO$ holds. In particular, by assuming that $O$ is the origin of $\mathbb{R}^2$ we have $H=A+B+C$ and
$$ HA^2+HB^2+HC^2 = \|B+C\|^2+\|A+C\|^2 +\|A+B\|^2 $$
where the polarization formula ensures
$$ \|B+C\|^2 = 2\|B\|^2+2\|C\|^2-\|B-C\|^2. $$
Thus the claim is proved, since $\|A\|=\|B\|=\|C\|=R$ and $\|B-C\|=BC$.
Of course $\|X-Y\|$ stands for the Euclidean distance between $X$ and $Y$, such that the polarization formula can be read as "in a parallelogram, the sum of the squared lengths of the sides equals the sum of the squared lengths of the diagonals".

You can also use this diagram, where segments with the same color have equal lengths, and recall that $H$ is the circumcenter of the anticomplementary triangle.
A: You may want to prove instead that $$AH^2+BC^2=4R^2.$$ This can be done by donsidering the parallelogram $AHBX$. Try to prove that $CX$ is the diameter of the circumcircle of triangle $ABC$.
A: hint: The following facts ( need proofs to get upvote...) can be used toward the proof:
1)  $r_a^2 + r_b^2+ r_c^2 +r^2 = AH^2 + BH^2 + CH^2 + 4R^2$
2)  $OH^2 = 9R^2 - AB^2 - BC^2 - AC^2$
3)  $r_a + r_b + r_c + r = AH + BH + CH + 2R$
4)  $r_a + r_b + r_c = 4R+r$
You may use these identities in your proof. 
Reference:  Coxeter and Greitzer Geometry Revisited by MAA
