2. Let $H$ be the orthocenter in an acute-angled triangle $ABC$ and $\overline{AB}=\overline{CH}$. Determine the measure of the angle at vertex $C$.
Solution. Angles $DAB$ and $DCH$ are equal as angles with normal sides, so the right triangles $\mathrm{ABD}$ and CHD are congruent. Therefore, $\overline{AD}=\overline{CD}$, making the right triangle $ADC$ isosceles, from which the angle at vertex $C$ is $45^{\circ}$.

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