Point $O$ lies in the interior of $\Delta ABC$. The bisectors of $\angle AOB, \angle BOC$ and $\angle COA$ intersect $\overline{AB}, \overline{BC}$ and $\overline{CA}$ at $D, E$ and $F$ respectively. Prove that $AD \times BE \times CF = DB \times EC \times FA$.

(A) In $\Delta AOB$,$OD$ is the angle bisector of $\angle AOB$. According to the Angle Bisector Theorem,the ratio of the segments of the side opposite to the angle is equal to the ratio of the other two sides of the triangle.
Therefore,$\frac{AD}{DB} = \frac{OA}{OB}$.
Similarly,in $\Delta BOC$,$OE$ is the angle bisector of $\angle BOC$. Thus,$\frac{BE}{EC} = \frac{OB}{OC}$.
In $\Delta COA$,$OF$ is the angle bisector of $\angle COA$. Thus,$\frac{CF}{FA} = \frac{OC}{OA}$.
Multiplying these three equations together:
$\frac{AD}{DB} \times \frac{BE}{EC} \times \frac{CF}{FA} = \frac{OA}{OB} \times \frac{OB}{OC} \times \frac{OC}{OA}$.
$\frac{AD \times BE \times CF}{DB \times EC \times FA} = 1$.
Therefore,$AD \times BE \times CF = DB \times EC \times FA$.