[The ratio of the areas of triangles with a common base or common height] It's complicated.

In triangle $ABC$, the sides $BC=a$, $AC=b$, $AB=c$, and the area $S$ are known. The bisectors $BM$ and $CN$ intersect at point $O$. Find the area of triangle $BOC$.
The bisector of a triangle divides its side into segments proportional to the other two sides, i.e.,

\[
\frac{B N}{A N}=\frac{B C}{A C}=\frac{a}{b}, \text{ and since } A B=c, \text{ then }
\]

\[
B N=A B \cdot \frac{B N}{A B}=c \cdot \frac{a}{a+b}=\frac{a c}{a+b}
\]

Therefore,

\[
S_{\triangle A N C}=\frac{B N}{A B} S_{\triangle A B C}=\frac{\frac{a c}{a+b}}{c} S=\frac{a}{a+b} S
\]

$B O$ is the bisector of triangle $B N C$, so

\[
\frac{C O}{O N}=\frac{B C}{B N}=\frac{a}{\frac{a c}{a+b}}=\frac{a+b}{c}, \frac{C O}{C N}=\frac{C O}{C O+O N}=\frac{a+b}{a+b+c}
\]

Therefore,

\[
S_{\triangle B O C}=\frac{C O}{C N} S_{\triangle M M C}=\frac{a+b}{a+b+c} \cdot \frac{a}{a+b} S=\frac{a S}{a+b+c}
\]

## Answer

\[
\frac{a S}{a+b+c}
\]