Let $O$ and $G$ be the circumcenter and centroid of $ABC$ respectively and let $K$ be the midpoint of $BC$ (Fig. 33).

We know that $AG = 2GK$. Also we know that $H$, $G$ and $O$ are collinear with $HG = 2GO$ (Euler line). So triangles $AHG$ and $KOG$ are similar with scale factor $2$ (by $2$ proportional sides and an equal angle between them). Therefore $AH = 2KO$.

Now the Pythagorean theorem in triangle $KOB$ yields $KO^2 + KB^2 = OB^2 = R^2$ and $AH^2 + BC^2 = (2KO)^2 + (2KB)^2 = 4(KO^2 + KB^2) = 4R^2$.

Let $B'$ be the other end of the diameter to the circumcircle of $ABC$ drawn from $B$ (Fig. 34). Since $AB'$ and $AB$ are perpendicular and $CH$ and $AB$ are perpendicular, the lines $AB'$ and $CH$ are parallel. Similarly, we see that $CB'$ and $AH$ are parallel, meaning that $AHCB'$ is a parallelogram. Thus $CB' = AH$.

Now the Pythagorean theorem in triangle $BCB'$ yields $CB'^2 + BC^2 = BB'^2$. As $CB' = AH$ and $BB' = 2R$, the desired result follows.