In $\triangle ABC$ , the internal bisector of the angle $\angle A$ meets $BC$ at $D$. If $AB=4$, $AC=3$ and $\angle A=60^{\circ}$, then the length of $AD$ is
$a.)\ 2\sqrt3\\
\color{green}{b.)\ \dfrac{12\sqrt3}{7}}\\
c.)\ \dfrac{15\sqrt3}{8}\\
d.)\ \dfrac{6\sqrt3}{7}\\$
With the help of cosine rule i found $BC$
Then with the relation $\dfrac{BD}{DC}=\dfrac{AB}{AC}=\dfrac43$,
I found $BD$ then again applying cosine rule in $\triangle ABD$ i found $AD$.
But i would like to know it there is another short simple way.
I have studied maths up to $12th$ grade.
Best Answer
Considering areas of triangles gives you $$\text{Area($\triangle{ABC}$)}=\text{Area($\triangle{ABD}$)}+\text{Area($\triangle{ACD}$)},$$ i.e. $$\frac 12\times 4\times 3\times\sin60^\circ=\frac 12\times4\times AD\times\sin30^\circ+\frac 12\times 3\times AD\times \sin30^\circ.$$