3 [Mutual arrangement of altitudes, medians, bisectors, etc.]

In triangle $ABC$, the bisector drawn from vertex $A$, the altitude drawn from vertex $B$, and the perpendicular bisector of side $AB$ intersect at one point. Find the angle at vertex $A$.
Let $\angle A=2 \alpha, M$ - the intersection point of the three lines specified in the condition. Since point $M$ lies on the perpendicular bisector of segment $A B$, then

$\angle A B M=\angle B A M=\alpha$.

From the right triangle $A B H$ we get $\alpha+2 \alpha=90^{\circ}$, that is $\alpha=30^{\circ}$. Therefore, $\angle B A C=60^{\circ}$.

## Answer

$60^{\circ}$.