Hello,

$\displaystyle \triangle$ABC is an acute triangle and H is the triangle's orthocenter.

Express the ratio of $\displaystyle \triangle$ABH's area to $\displaystyle \triangle$ABC's area through the triangle's angles.

(No sketch is given)

According to the answer key, the answer is supposed to be:

$\displaystyle \frac{\triangle ABH}{\triangle ABC} = \cot\alpha * \cot\beta$

Any help?

Thanks in advance.