Question

If the cotangents of half the angles of a triangle are in A.P. then prove that the sides are in A.P.

Answer 1

we know

$\cot \frac{A}{2}=\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}=\frac{s(s-a)}{\Delta }$

Similarly, $\cot \frac{B}{2}=\frac{s(s-b)}{\Delta }$ and $\cot \frac{C}{2}=\frac{s(s-c)}{\Delta }$

Given $\cot \frac{A}{2},\cot \frac{B}{2}$ and $\cot \frac{C}{2}$ are in A.P.
$\Rightarrow \frac{s(s-a)}{\Delta },\frac{s(s-b)}{\Delta }$ and $\frac{s(s-c)}{\Delta }$ are in A.P.

$\Rightarrow s-a,s-b$ and $s-c$ are in A.P.

$\Rightarrow a,b,$ and c are in A.P.