5.1. (14 points) In an acute-angled triangle $A B C$, angle $A$ is equal to $35^{\circ}$, segments $B B_{1}$ and $C C_{1}$ are altitudes, points $B_{2}$ and $C_{2}$ are the midpoints of sides $A C$ and $A B$ respectively. Lines $B_{1} C_{2}$ and $C_{1} B_{2}$ intersect at point $K$. Find the measure (in degrees) of angle $B_{1} K B_{2}$.
Answer: 75.

![](https://cdn.mathpix.com/cropped/2024_05_06_8ed28a093e6868403444g-2.jpg?height=408&width=671&top_left_y=2003&top_left_x=1252)

Solution. Note that angles $B$ and $C$ of triangle $ABC$ are greater than $\angle A=35^{\circ}$ (otherwise it would be an obtuse triangle), so point $C_{1}$ lies on side $AB$ between points $B$ and $C_{2}$, and point $B_{1}$ lies on side $AC$ between points $C$ and $B_{2}$. Therefore, the point $K$ of intersection of lines $B_{1} C_{2}$ and $C_{1} B_{2}$ lies inside the triangle. Since $C_{1} B_{2}$ is the median of the right triangle $C C_{1} A$, triangle $C_{1} B_{2} A$ is isosceles. Therefore, $\angle A B_{1} C_{2}=35^{\circ}$. Similarly, we get $\angle A C_{1} B_{2}=35^{\circ}$, from which $\angle A B_{2} C_{1}=180^{\circ}-2 \cdot 35^{\circ}=110^{\circ}$. Then $\angle B_{1} K B_{2}=\angle A B_{2} K-\angle A B_{1} K=110^{\circ}-35^{\circ}=75^{\circ}$.