[The ratio in which the bisector divides the side.]

In triangle $ABC$, it is known that $AB=14, BC=6, AC=10$. The bisectors $BD$ and $CE$ intersect at point $O$. Find $OD$.

#
The bisector of a triangle divides its side into segments proportional to the other two sides. Apply this theorem to triangles $A B C$ and $B C D$.

## Solution

By the property of the bisector of a triangle

$$
\frac{C D}{A D}=\frac{B C}{A B}=\frac{6}{14}=\frac{3}{7}
$$

From this, it follows that $C D=3$. Similarly, we find that

$$
\frac{B O}{O D}=\frac{B C}{C D}=\frac{6}{3}=2
$$

By the cosine rule in triangle $A B C$, we find that

$$
\cos \angle A C B=\frac{A C^{2}+B C^{2}-A B^{2}}{2 \cdot A C \cdot B C}=\frac{100+36-196}{2 \cdot 10 \cdot 6}=-\frac{60}{120^{\circ}}=-\frac{1}{2}
$$

Therefore, $\angle A C B=120^{\circ}$. Then, by the cosine rule in triangle $B C D$, we find that

$$
B D^{2}=B C^{2}+C D^{2}-2 \cdot B C \cdot C D \cdot \cos 120^{\circ}=36+9+18=63
$$

Therefore,

$$
B D=\sqrt{63}=3 \sqrt{7}, \mathrm{a} O D=\frac{1}{3} B D=\frac{1}{3} \cdot 3 \sqrt{7}=\sqrt{7}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_aa2960437c2efca8c53fg-36.jpg?height=286&width=693&top_left_y=111&top_left_x=683)

## Answer

$\sqrt{7}$.