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 ...


2 Answers 2


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.

  • $\begingroup$ You have taken all the three points of ABC on $y$ axis,how is this possible? $\endgroup$
    – diya
    Aug 6, 2015 at 15:14
  • $\begingroup$ @diya That's a typo. Thanks. $\endgroup$ Aug 6, 2015 at 15:16
  • $\begingroup$ Sir,i understood your method,but i want to ask if there is some other geometrical method to prove this formula.Can we do it without supposing the coordinate of the vertices?Thanks. $\endgroup$
    – diya
    Aug 6, 2015 at 15:39
  • $\begingroup$ You can just note that by the property of scaling of axis, rotation of axis, and translation of axis. This is just good enough to work with any other case. $\endgroup$
    – Someone
    Aug 6, 2015 at 16:35
  • $\begingroup$ @diya You can translate it into trigonometric language. But the equality to be proved is so complicated that without coordinates the computation would be difficult. $\endgroup$ Aug 7, 2015 at 0:49

Let $a$ be the length of side BC coinciding with the x-axis such that vertex B is at origin $(0, 0)$ & vertex C is at $(a, 0)$ then vertex A will at $\left(\frac{a\tan C}{\tan B+\tan C}, \frac{a\tan B\tan C}{\tan B+\tan C}\right)$ Hence, the ortho-center will be $$H\equiv \left(\frac{0+a+\frac{a\tan C}{\tan B+\tan C}}{3}, \frac{0+0+\frac{a\tan B\tan C}{\tan B+\tan C}}{3} \right)\equiv \left(\frac{a(\tan B+2\tan C)}{3(\tan B+\tan C)}, \frac{a\tan B\tan C}{3(\tan B+\tan C)} \right)$$ & the circumscribed center say $D$ can be calculated as $$D\equiv\left(\frac{a}{2}, \frac{a(\tan B\tan C-1)}{2(\tan B+\tan C)} \right)$$

Hence, the slope of the line HD joining $H$ & $D$ with the BC (x-axis) is given as $$m=\frac{y_2-y_1}{x_2-x_1}$$ $$=\frac{\frac{a(\tan B\tan C-1)}{2(\tan B+\tan C)}-\frac{a\tan B\tan C}{3(\tan B+\tan C)}}{\frac{a}{2}-\frac{a(\tan B+2\tan C)}{3(\tan B+\tan C)}}$$ $$=\frac{3\tan B\tan C-3-2\tan B\tan C}{3\tan B+3\tan C-2\tan B-4\tan C}$$ $$=\frac{\tan B\tan C-3}{\tan B-\tan C}$$ Hence the angle of the line joining H & D with side BC is given as $$\tan \theta=|m|$$ $$\implies \theta=\tan^{-1}\left|\frac{\tan B\tan C-3}{\tan B-\tan C}\right|$$ or $$\bbox [5px, border:2px solid #C0A000]{\color{red}{\theta=\tan^{-1}\left|\frac{3-\tan B\tan C}{\tan B-\tan C}\right|}}$$


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