Prove that $AH^2+BC^2=4AO^2$

Question

Prove that $AH^2+BC^2=4AO^2$, where $O$ is the circumcentre and $H$ is the orthocentre of the triangle $ABC$.

Answer 1

Hint: Let $BD$ be a diameter of the circumcircle of $ABC$. Show that $CD=AH$.

Answer 2

We know that $AH=2\cdot OD$. (where D is the midpoint of BC)
By using Pythagoras theorem we get:
$$\implies OD^2+BD^2=OB^2=OA^2$$ $$\implies \left(\frac{AH}{2}\right)^2+\left(\frac{BC}{2}\right)^2=OA^2$$ $$\implies AH^2+BC^2=4\cdot OA^2$$ (Hence Proved)

Answer 3

There is no loss of generality in assuming that the $O$ is at the origin.If $Rz_1, Rz_2, Rz_3$ are complex numbers representing $A,B,C$ respectively, then the orthocenter is $R(z_1+z_2+z_3)$. Thus \begin{align*} AH^2 + BC^2 = R^2(|z_2+z_3|^2 + |z_2-z_3|^2) = 2R^2(|z_2|^2+|z_3|^2) = 4R^2 = 4AO^2 \end{align*} since $OA = R$, the radius of the circum circle.