8. Given triangle $A B C . \angle A=\alpha, \angle B=\beta$. Lines $O_{1} O_{2}, O_{2} O_{3}, O_{1} O_{3}$ are the bisectors of the external angles of triangle $A B C$, as shown in the figure. Point $\mathrm{O}$ is the center of the inscribed circle of triangle $A B C$. Find the angle between the lines $O_{1} O_{2}$ and $O O_{3}$.

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# Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_de39ee82fbf77ee5c736g-6.jpg?height=629&width=608&top_left_y=1830&top_left_x=838)

We will prove that the bisectors of two external angles and one internal angle intersect at one point. Let \( O_{1} G, O_{1} H, O_{1} F \) be the perpendiculars to \( B C, A C \) and \( A B \) respectively. Then triangles \( A H O_{1} \) and \( B F O_{1} \), \( B F O_{1} \) and \( B G O_{1} \) are right triangles and are equal by the hypotenuse and acute angle. From the equality of the triangles, it follows that \( H O_{1} = G O_{1} \). Therefore, \( O_{1} \) is equidistant from the sides of angle \( C \), which means \( C O_{1} \) is the bisector of angle \( C \). Similarly, it can be proven that \( A O_{2} \) and \( B O_{3} \) are the bisectors of angles \( A \) and \( B \). The point \( O \) is the point of intersection of the bisectors of triangle \( \mathrm{ABC} \). The angle \( O_{1} A O_{2} \) is a right angle because it is formed by the bisectors of adjacent angles. Similarly, the angles \( O_{1} B O_{3} \) and \( O_{1} C O_{2} \) are right angles, and therefore, \( O \) is the point of intersection of the altitudes of triangle \( O_{1} O_{2} O_{3} \). Thus, \( O_{3} O \) is perpendicular to \( \mathrm{O}_{1} \mathrm{O}_{2} \).

Answer: \( 90^{\circ} \).