46.2. In an isosceles triangle, the altitudes intersect on the inscribed circle. Find the angles of the triangle.

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\text { (9-10 grades) }
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46.2. Let in an isosceles triangle \(ABC (AB = BC)\) the altitudes \(AK, BD\), and \(CM\) intersect at point \(P\), which lies on the inscribed circle with center \(O\) (Fig. 22). Let \(\angle BAC = \alpha\).

![](https://cdn.mathpix.com/cropped/2024_05_21_f0d089db3d527e92a992g-35.jpg?height=497&width=648&top_left_y=565&top_left_x=704)

This angle is common to the right triangles \(ABD\) and \(ACM\). Therefore, \(\angle ABD = \angle ACM = 90^\circ - \alpha\). Since triangle \(APC\) is isosceles, the angle \(\angle PAC\) is also equal to \(90^\circ - \alpha\). Then, in the right triangle \(APD\), the angle \(\angle APD\) is equal to \(\alpha\). Therefore, \(AD : DP = \tan \alpha\). On the other hand, considering that \(AO\) is the bisector of angle \(BAD\), we get:

\[
AD : DP = AD : (2 \cdot OD) = (AD : OD) : 2 = \frac{1}{2} \cot \frac{\alpha}{2}
\]

We have the equation:

\[
\tan \alpha = \frac{1}{2} \cot \frac{\alpha}{2}
\]

i.e., \(\frac{2 \tan \frac{\alpha}{2}}{1 - \tan^2 \frac{\alpha}{2}} = \frac{1}{2 \tan \frac{\alpha}{2}} ; 5 \tan^2 \frac{\alpha}{2} = 1 ; \quad \alpha = 2 \arctan \frac{1}{\sqrt{5}}\).

Thus, \(\angle BAC = \angle BCA = 2 \arctan \frac{1}{\sqrt{5}}, \angle ABC = \pi - 4 \arctan \frac{1}{\sqrt{5}}\).