[ The ratio in which the bisector divides the side. ] [ The ratio of the areas of triangles with a common angle ]

In triangle $ABC$, the bisectors $AD$ and $BE$ intersect at point $O$. Find the ratio of the area of triangle $ABC$ to the area of quadrilateral $ODCE$, given that $BC=a$, $AC=b$, and $AB=c$.
$\frac{S_{\triangle B O D}}{S_{\delta B E C}}=\frac{B D}{B C} \cdot \frac{B O}{B E}$.

## Solution

Let $S_{\Delta \mathrm{ABC}}=S$. By the property of the angle bisector of a triangle,

$$
\frac{B D}{C B}=\frac{c}{b+c}, A E=\frac{b c}{a+c}, \frac{B O}{B E}=\frac{a+c}{a+b+c}
$$

Then,

$$
S_{\triangle \mathrm{BOD}}=\frac{B D}{C B} \cdot \frac{B O}{B E} S_{\triangle \mathrm{BEC}}=\frac{c}{b+c} \cdot \frac{a+c}{a+b+c} \cdot \frac{a}{a+c} \cdot S=\frac{a c S}{(b+c)(a+b+c)} .
$$

Therefore,

$$
\begin{gathered}
S_{\mathrm{ODCE}}=S_{\triangle \mathrm{BEC}}-S_{\triangle \mathrm{BOD}}=\frac{a S}{a+c}-\frac{a c S}{(b+c)(a+b+c)}= \\
=\frac{a b(a+b+2 c) S}{(a+c)(b+c)(a+b+c)} .
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5827e7e7a03e1b3b311bg-22.jpg?height=383&width=732&top_left_y=789&top_left_x=663)

Answer $\frac{(a+c)(b+c)(a+b+c)}{a b(a+b+2 c)}$.

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