Problem 3. Given an acute triangle $ABC$. Its altitudes $AE$ and $CD$ intersect at point $H$, and $\overline{AB} = \overline{CH}$. Calculate the angle $\measuredangle ACB$.
Solution. Since $\measuredangle E A B=\measuredangle B C D$ (angles with perpendicular sides), triangles $A B E$ and $C H E$ are congruent (by the ASA criterion). Now $\overline{C E}=\overline{A E}$, meaning triangle $A E C$ is an isosceles right triangle. Therefore,

$$
\measuredangle B A C=\measuredangle E C A=45^{\circ}
$$

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