# The heights of the acute-angled triangle ABC, drawn from points B and C, continued to intersect with the circumscribed circle

The heights of the acute-angled triangle ABC, drawn from points B and C, continued to intersect with the circumscribed circle at points B1 and C1. It turned out that the segment B1C1 passes through the center of the circumscribed circle. Find the angle of the BAC.

∠BAC is an inscribed angle and rests on a small arc CB.
Draw a segment CB1, ∠CB1B is also inscribed and is based on the same arc, therefore, ∠BAC = ∠CB1B.
B1C1 is the diameter of the circle, as it passes through its center. Therefore, B1C1 divides the circle into two arcs of 180 °
∠B1CC1 is also inscribed and relies on an arc of 180 °, according to the inscribed angle theorem ∠B1CC1 = 180 ° / 2 = 90 °.
We will designate three more points, as shown in the figure below:
Points E and F are the points of intersection of the heights and sides of the triangle ABC, G is the point of intersection of heights.
Consider the triangles B1CG and BFG.
∠CGB1 = ∠BGF (since they are vertical).
∠B1CG = ∠BFG (since they are both straight).
Therefore, by the theorem on the sum of the angles of a triangle, ∠СB1G = ∠GBF
Therefore, ∠GBF is also equal to ∠BAC
Consider the triangle AEB.
∠AEB = 90 ° (since BE is the height).
∠BAC = ∠GBF
Then, using the theorem on the sum of the angles of a triangle, we obtain that each of the angles BAC and GBF is equal to 45 °.