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

$a.)\ 2\sqrt3\\ \color{green}{b.)\ \dfrac{12\sqrt3}{7}}\\ c.)\ \dfrac{15\sqrt3}{8}\\ d.)\ \dfrac{6\sqrt3}{7}\\$

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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.


Considering areas of triangles gives you $$\text{Area($\triangle{ABC}$)}=\text{Area($\triangle{ABD}$)}+\text{Area($\triangle{ACD}$)},$$ i.e. $$\frac 12\times 4\times 3\times\sin60^\circ=\frac 12\times4\times AD\times\sin30^\circ+\frac 12\times 3\times AD\times \sin30^\circ.$$

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    $\begingroup$ OMG! brilliant!! $\endgroup$ – R K May 22 '15 at 22:30

Let the length of AD be $x$ & $\angle ADC=\theta$

Apply law of sine in $\Delta ABD$ $$\frac{\sin(\pi-\theta)}{AB}=\frac{\sin 30^o}{BD} \implies BD=\frac{4}{2\sin\theta}=\frac{2}{\sin\theta}$$ Apply law of cosine in $\Delta ABD$ $$BD=\sqrt{4^2+x^2-2(4)(x)\cos 30^o}\implies \frac{2}{\sin\theta}=\sqrt{x^2-4x\sqrt{3}+16}\tag 1$$ Similarly, apply law of sine in $\Delta ADC$ $$\frac{\sin\theta}{AC}=\frac{\sin 30^o}{DC} \implies DC=\frac{3}{2\sin\theta}$$ Now, apply law of cosine in $\Delta ADC$ $$DC=\sqrt{3^2+x^2-2(3)(x)\cos 30^o}\implies \frac{3}{2\sin\theta}=\sqrt{x^2-3x\sqrt{3}+9}\tag 2$$ Now, dividing the eq (1) by (2), we get $$\frac{4}{3}=\frac{\sqrt{x^2-4x\sqrt{3}+16}}{\sqrt{x^2-3x\sqrt{3}+9}}\implies 7x^2-12\sqrt{3}x=0 \implies x=\frac{12\sqrt{3}}{7} \space (x>0)$$ Hence, the length of AD is $\frac{12\sqrt{3}}{7}$ Thus option (b) is correct.


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