2. Let the orthocenter $H$ of the isosceles triangle $ABC$ lie on the inscribed circle. Find the cosine of the angle $\alpha$ at the base $AB$ of this triangle.
Solution. Let $O$ be the center of the inscribed circle in the isosceles triangle $ABC$ (see diagram). Then we have $\measuredangle OAD = \frac{\alpha}{2}, \measuredangle AHD = \alpha$. From the right triangles $AHD$ and $AOD$, we get $\overline{HD} = \overline{AD} \operatorname{ctg} \alpha$ and $\overline{OD} = \overline{AD} \operatorname{tg} \frac{\alpha}{2}$. Since $\overline{HD} = 2 \overline{OD}$, we get $\operatorname{ctg} \alpha = 2 \operatorname{tg} \frac{\alpha}{2}$, from which we also get $\cos \alpha = \frac{2}{3}$.

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