Folklore

In an acute-angled triangle $A B C$, the bisector $A N$, the altitude $B H$, and the line perpendicular to side $A B$ and passing through its midpoint intersect at one point. Find the angle $BAC$.

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Let $P$ be the intersection point of $A N$ and $B H$ (see the figure). Since the perpendicular bisector of side $A B$ passes through point $P$, we have $P A = P B$, which means triangle $A P B$ is isosceles. Therefore, $\angle A B P = \angle B A P = \angle P A H$, since $A P$ is the angle bisector of $\angle B A H$. Moreover, $\angle A H B = 90^{\circ}$, so the sum of these three equal angles is $90^{\circ}$, meaning each of them is $30^{\circ}$. Thus, $\angle B A C = 60^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ae6ec2c916c776dd1759g-23.jpg?height=412&width=598&top_left_y=1252&top_left_x=730)

## Answer

$60^{\circ}$.