Question

In $△ABC$, if AD is the bisector of $\angle A$, prove that

$\frac{Area\left(△ABD\right)}{Area\left(△ACD\right)}=\frac{AB}{AC}$

Answer 1

In $△ABC$, AD is the bisector of $\angle A$.

$\therefore$ $\frac{AB}{AC}=\frac{BD}{DC}$.......(i)

From A, draw AL $\perp$ BC.

$\therefore$ $\frac{Area\left(△ABD\right)}{Area\left(△ACD\right)}=\frac{\left(1/2\right)BD.AL}{\left(1/2\right)DC.AL}$

$\Rightarrow$ $\frac{Area\left(△ABD\right)}{Area\left(△ACD\right)}=\frac{BD}{DC}$

$\Rightarrow$ $\frac{Area\left(△ABD\right)}{Area\left(△ACD\right)}=\frac{AB}{AC}$ ......... $\left[From\right(i\left)\right]$ $\left[Henceproved\right]$