In $\Delta ABC, $ $O$ is the circumcentre and $H$ is the orthocentre. Then prove that $AH^2 +BC^2 =4AO^2.$

Question

In $\Delta ABC, $ $O$ is the circumcentre and $H$ is the orthocentre. Then prove that $AH^2 +BC^2 =4AO^2.$

I am unable to solve this problem. Any help will be appreciated.

Answer 1

Let the circum center be taken as the origin and $z_1, z_2, z_3$ the complex numbers representing the vertices $A,B,C$ respectively. It is not difficult to show that the orthocenter $H$ is represented by the complex number $z_1+z_2+z_3$. Thus \begin{align*} AH^2 + BC^2 &= |z_1+z_2+z_3 - z_1|^2 + |z_2-z_3|^2 \\ &= |z_2+z_3|^2 +|z_2-z_3|^2\\ &= 2(|z_2|^2 +|z_3|^2) \\ &= 4R^2 \end{align*} since $|z_2| = |z_3| = R$.

Answer 2

It is well known that $AH=2R\cos A$. We already know that $AO=R$. So, we are trying to prove the following statement: $$4R^2\cos^2 A+BC^2\stackrel{?}{=}4R^2$$

Which is equivalent to this statement: $$BC^2\stackrel{?}{=}4R^2\left(1-\cos^2 A\right)$$ Note that $1-\cos^2 A=\sin^2 A$: $$BC^2\stackrel{?}{=}4R^2\sin^2 A$$

Take the square root of both sides (we know $BC$ must be positive): $$BC\stackrel{?}{=}2R\sin A$$

Divide by $\sin A$: $$\frac{BC}{\sin A}\stackrel{?}{=}2R$$ This is true because of the extended law of sines.

Answer 3

$(C)$ is the circumcircle. Let $$\text{AO}\cap (C)=D$$ $$\text{AH}\cap (C)=E$$ $$\text{AE}\cap \text{BC}=I$$ $$\text{HD}\cap \text{BC}=N$$ $HIN$ and $HED$ are right triangles in which $I$ is the midpoint of $HE$ implies $N$ is the midpoint of $HD$. This with the fact that $O$ is the midpoint of $AD$ implies $ON$ is parallel to $AH$ and is perpendicular to $BC$. Also, $ON=AH/2$.

So $$\text{ON}^2=\text{OB}^2-(\text{BC}/2)^2=\text{OA}^2-(\text{BC}/2)^2=(\text{AH}/2)^2$$

Therefore $$4\text{OA}^2=\text{BC}^2+\text{AH}^2$$