In $\vartriangle ABC$, if $cot \frac{A}{2},\ cot \frac{B}{2},\ cot \frac{C}{2}$ are in A.P. Then show that $a,\ b,\ c$ are in A.P.  

Given: In $\vartriangle ABC$, $cot \frac{A}{2},\ cot \frac{B}{2},\ cot \frac{C}{2}$  are in A.P.

To do: To show that $a,\ b,\ and\ c$ are in A.P.

$cot\frac{A}{2}=\frac{S−a}{r}$

$cot\frac{B}{2}=\frac{S−b}{r}$

$cot\frac{C}{2}=\frac{S−c}{r}$

​

 $S=\frac{a+b+c}{2}$

​

$r= radius$

$\frac{cot( \frac{A}{2})}{S-a}=\frac{cot( \frac{B}{2})}{S-b}=\frac{cot( \frac{C}{2})}{S-c}$

​  

As $cot\frac{A}{2};cot\frac{B}{2};cot\frac{C}{2}$ are in A.P, 

$\therefore cot\frac{B}{2}=\frac{[cot\frac{A}{2}+cot\frac{C}{2}]}{2}$

So $\frac{[cot\frac{A}{2}+cot\frac{C}{2}]}{2}=\frac{[( \frac{( S−a)}{(S−b)})cot( \frac{B}{2})+( \frac{( S−c)}{( S−b)})cot\frac{B}{2}]}{2}$

​

$=\frac{1}{2}.( \frac{( 2S−a−c)}{S−b})cot\frac{B}{2}$

$=\frac{1}{2}.\frac{( a+2b+c)}{( a+c)}cot\frac{B}{2}$

$=\frac{1}{2}.( \frac{( 2a+2c)}{a+c})cot\frac{B}{2}$  [Since $2b=a+c$]

$=cot\frac{B}{2}$

$\therefore a,\ b,\ c$ are in A.P.