3. Given an acute-angled $\triangle ABC$. The angle bisector of $\measuredangle BAC$, the perpendicular bisector of side $AC$, and the altitude from point $C$ intersect at one point. Determine the angle $\measuredangle BAC$!
Solution. Let $\angle B A C=\alpha$ and let $S$ be the intersection. Since $S$ lies on the perpendicular bisector of side $A C$, we have $\overline{A S}=\overline{S C}$, which means triangle $\triangle$ ASC is isosceles. Therefore,

$$
\measuredangle S A C=\measuredangle A C S=\frac{\alpha}{2} .
$$

On the other hand, $\triangle A E C$ is a right triangle, so,

$$
\begin{aligned}
\measuredangle A C S & =180^{\circ}-\alpha-90^{\circ} \\
& =90^{\circ}-\alpha
\end{aligned}
$$

Finally, from (1) and (2) we get $\frac{\alpha}{2}=90^{\circ}-\alpha$, i.e., $\alpha=60^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_bf69b18cff2f2eea848fg-2.jpg?height=361&width=716&top_left_y=257&top_left_x=781)