Q:

Line joining the orthocenter to the circumcenter of a triangle ABC is inclined to BC at an angle $\tan^{-1}(\frac{3-\tan B\tan C}{\tan B-\tan C})$

Show that the line joining the orthocenter to the circumscribed center of a triangle ABC is inclined to BC at an angle $\tan^{-1}\left(\frac{3-\tan B\tan C}{\tan B-\tan C}\right)$
I let the foot of perpendicular from A,B,C to opposite sides is D,E,F.Then
$$\tan B=\frac{AD}{BD},\tan C=\frac{AD}{CD}$$
$$\frac{3-\tan B\tan C}{\tan B-\tan C}=\frac{3-\frac{AD}{BD}\frac{AD}{CD}}{\frac{AD}{BD}-\frac{AD}{CD}}$$
I think this way i cannot get answer.Please help me getting the desired proof.I am stuck ...

A:

Suppose $D=(0,0), A=(0,1),B=(b,0),C=(c,0)$. 
Then the orthocenter $H=(0,-bc)$ and the circumcenter $O=(\frac{b+c}2,\frac{bc+1}2)$ after simple computation. The slope of $OH$ is $\frac{3bc+1}{b+c}$. 
Note that $\tan B=-\frac1b$ and $\tan C=\frac1c$. The proof is completed.

