5. The heights of an acute-angled triangle $ABC$, drawn from points $B$ and $C$, were extended to intersect the circumscribed circle at points $B_{1}$ and $C_{1}$. It turned out that the segment $B_{1} C_{1}$ passes through the center of the circumscribed circle. Find the angle $BAC$.
# Solution

Since $B_{1} C_{1}$ is the diameter of the circle, $\angle B_{1} B C_{1}=\angle B_{1} C C_{1}=90^{\circ}$, therefore, $B C_{1} \| A C$ and $C B_{1} \| A B$ (see figure). Since $B C_{1} \| A C$, then $\angle C_{1} B A=\angle A=\alpha$. Similarly, $\angle B_{1} C A=\angle A=\alpha$. The degree measure of the arc $B_{1} C_{1}$ is $180^{\circ}$, so the sum of the inscribed angles $C_{1} B A$ and $B_{1} C A$ is $90^{\circ}$, which means $\angle A=45^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_4bea69d95f48fa3f775dg-2.jpg?height=294&width=349&top_left_y=2172&top_left_x=268)

Answer: $45^{\circ}$.

Remarks: The answer can also be obtained by proving that triangle $C D B_{1}$ is an isosceles right triangle.