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  4. In a triangle ABC, the angle bisector BD of angle B intersects AC in D. Suppose BC =2, CD =1 and BD =(3/√2). The perimeter of the triangle ABC is

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Q. In a triangle $ABC$, the angle bisector BD of $\angle B$ intersects AC in D. Suppose $BC =2, CD =1$ and $BD =\frac{3}{\sqrt{2}}$. The perimeter of the triangle $ABC$ is

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Solution:

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$\frac{2 ac }{ a + c } \cos \frac{ B }{2}=\frac{3}{\sqrt{2}}$
$\Rightarrow \frac{4 c}{2+c}\left[\frac{4+\frac{9}{2}-1}{2 \times 2 \times \frac{3}{\sqrt{2}}}\right]=\frac{3}{\sqrt{2}}$
$\Rightarrow c=3$
Now, $\frac{c}{a}=\frac{A D}{D C} \Rightarrow A D=\frac{3}{2}$
$\Rightarrow b =\frac{5}{2}$
$\Rightarrow \text { Perimeter }=\frac{15}{2}$