36.3. In triangle $A B C$, the bisectors $A A_{1}$ and $B B_{1}$ intersect at point $O$. Find the ratio $A A_{1}: O A_{1}$, if $A B=6, B C=5$, $C A=4$.
36.3. First, we will prove that the bisector of an angle in a triangle divides the opposite side into segments proportional to the adjacent sides. We will use the bisector $B B_{1}$ (Fig. 28).

![](https://cdn.mathpix.com/cropped/2024_05_21_f0d089db3d527e92a992g-42.jpg?height=365&width=694&top_left_y=640&top_left_x=681)

We need to prove that $A B: C B = A B_{1}: C B_{1}$. On the line $B B_{1}$, take a point $B_{2}$ such that $A B = A B_{2}$. We obtain similar triangles $A B_{1} B_{2}$ and $C B_{1} B$, since $\angle A B_{1} B_{2} = \angle C B_{1} B$ and $\angle A B_{2} B_{1} = \angle C B B_{1}$. From the similarity of these triangles, we have: $A B_{2}: B C = A B_{1}: C B_{1}$. But $A B_{2} = A B$. Therefore, $A B: B C = A B_{1}: C B_{1}$, which is what we needed to prove. Similarly, for the bisector $A A_{1}$, we get: $A B: A C = B A_{1}: C A_{1}$. But $A B = 6$ and $A C = 4$. Then $B A_{1}: C A_{1} = 3: 2$. Since $B A_{1} + C A_{1} = B C = 5$, we have $B A_{1} = 3$. Now, from triangle $A B A_{1}$ for the bisector $B O$, we have: $A B: B A_{1} = A O: O A_{1}$, from which $A O: O A_{1} = 2: 1$. Therefore, $A A_{1}: O A_{1} = 3: 1$.