8,9 Find the angles of an isosceles triangle, knowing that the point of intersection of its heights lies on the inscribed circle.

#
Let $O$ - the center of the circle inscribed in triangle $ABC (AC = BC)$, $H$ - the point of intersection of the altitudes, $\angle CAB = \angle CBA = \alpha$, $K$ - the midpoint of $AB$. Then

$$
O K = A K \operatorname{tg} \frac{\frac{1}{2}}{2}, H K = A K \operatorname{tg}\left(90^{\circ} - \alpha\right) = A K \operatorname{ctg} \alpha
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_786127613a7b2f99b094g-27.jpg?height=74&width=1718&top_left_y=2673&top_left_x=16)

$$
2 t = \frac{1 - t^{2}}{2 t}
$$

## Answer

$\arccos \frac{2}{3}, \arccos \frac{2}{3}, \pi - 2 \arccos \frac{2}{3}$.