Question

In $\Delta ABC$ if $(a-b)(s-c)=(b-c)(s-a)$, then $r_1,r_2,r_3$ are in

• A. A.P.
• B. G.P.
• C. H.P.
• D. A.G.P.

Answer 1

The correct option is A A.P.

Let, $\Delta$ be the area of given triangle

Given,

$(a-b)\times (s-c)=(b-c)\times (s-a)$

$\Rightarrow$ $\left[\right(s-a)-(s-b\left)\right]\times (s-c)=\left[\right(s-b)-(s-c\left)\right]\times \left(\right(s-a)$

Dividing both sides by $\left[\right(s-a)\times (s-b)\times (s-c\left)\right]$, we get,

$\Rightarrow$ $[\frac{1}{(s-b)}-\frac{1}{(s-a)}]=[\frac{1}{(s-c)}-\frac{1}{(s-b)}]$

Multiplying both sides by $\Delta$, we get,

$\frac{\Delta }{(s-b)}-\frac{\Delta }{(s-a)}$ $=\frac{\Delta }{(s-c)}-\frac{\Delta }{(s-b)}$

Since we know that,

$r_1=\frac{\Delta }{(s-a)}$

$r_2=\frac{\Delta }{(s-b)}$

$r_3=\frac{\Delta }{(s-c)}$

Thus, $r_2-r_1=r_3-r_2$

$\Rightarrow$ $2\times r_2=r_1+r_3$

Hence, $r_1,r_2,r_3$ are in A.P.