Question

In given figure O is any point inside a triangle ABC. The bisector of $\angle AOB$, $\angle BOC$ and $\angle COA$ meet the sides AB, BC and CA in point D,E and F respectively. Show that
$AD\times BE\times CF=DB\times EC\times FA$

Answer 1

In $△AOD$, OD is the bisector of $\angle AOB$.

$\therefore$ $\frac{OA}{OB}=\frac{AD}{DB}$........(i)

In $△BOC$, OE is the bisector of $\angle BOC$.

$\therefore$ $\frac{OB}{OC}=\frac{BE}{EC}$......(ii)

In $△COA$, OF is the bisector of $\angle COA$.

$\therefore$ $\frac{OC}{OA}=\frac{CF}{FA}$.......(iii)

Multiplying the corresponding sides of (i), (ii) and (iii), we get

$\frac{OA}{OB}\times \frac{OB}{OC}\times \frac{OC}{OA}=\frac{AD}{DB}\times \frac{BE}{EC}\times \frac{CF}{FA}$

$\Rightarrow$ $1=\frac{AD}{DB}\times \frac{BE}{EC}\times \frac{CF}{FA}$

$\Rightarrow$ $DB\times EC\times FA=AD\times BE\times CF$

$\Rightarrow$ $AD\times BE\times CF=DB\times EC\times FA$ $\left[Henceproved\right]$