Let $O$, $H$ and $R$ be the circumcenter, orthocenter and circumradius of $\Delta ABC$. Let $BC=a, CA=B$ and $AB=c$. Then by Law of Sines, $$R=\frac{a}{2\sin A}=\frac{a}{2\sin 120^{\circ}}=\frac{a}{\sqrt{3}}.$$Now, by Law of Cosines, \begin{align*} a^2=b^2+c^2-2bc\cos A & = b^2+c^2-2bc\cos 120^{\circ} \\ & = b^2+c^2+bc. \\ \end{align*}Finally, \begin{align*} OH^2 = 9R^2-a^2-b^2-c^2 & = 2a^2-b^2-c^2 \\ & = b^2+c^2+2bc \\ & = (b+c)^2. \\ \end{align*}$$\therefore \boxed{OH=b+c}$$$QED.$