12.409 In an acute-angled triangle $ABC$, the angles are known. Find the ratio in which the orthocenter (the point of intersection of the altitudes) divides the altitude drawn from vertex $A$.
Solution.

Let $O$ be the orthocenter of the given $\triangle ABC$,

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-292.jpg?height=274&width=364&top_left_y=1326&top_left_x=105)

$AD$ - its altitude. Then $S_{OBD} = \frac{BD \cdot BO}{26} \cdot \sin (\angle OBD)$,

$$
S_{OBA} = \frac{AB \cdot BO}{2} \cdot \sin (\angle OBA)
$$

But $\angle OBD = 90^{\circ} - \angle C, \angle OBA = 90^{\circ} - \angle A \Rightarrow$

$\Rightarrow \frac{S_{OBA}}{S_{OBD}} = \frac{AB}{BD} \cdot \frac{\cos \angle A}{\cos \angle C} = \frac{\cos \angle A}{\cos \angle B \cdot \cos \angle C}$.

But, by Theorem 10, $\frac{S_{OBA}}{S_{OBD}} = \frac{AO}{OD}$.

Answer: $\frac{\cos \angle A}{\cos \angle B \cdot \cos \angle C}$.