The altitudes of the 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 $BAC$.

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Let $\angle B A C=\alpha$. Inscribed angles $B A C$ and $B B 1 C$ subtend the same arc, so $\angle$ $B B 1 C=\angle B A C=\alpha$. Point $C$ lies on the circle with diameter $B 1 C 1$, therefore $\angle B 1 C C 1=90^{\circ}$. Lines $B 1 C$ and $A B$ are parallel, as they are perpendicular to the same line $C C 1$, hence $\angle A C B 1=\angle B A C=\alpha$. Let segments $B B 1$ and $A C$ intersect at point $D$. In the right triangle $C B 1 D$, each of the acute angles is $\alpha$, therefore, $\alpha=45^{\circ}$.

## Answer

$45^{\circ}$.