In ∆ A B C , suppose the radius of the circle opposite to an angle A is denoted by r 1 , similarly r 2…

We know, $r_1=\frac{∆}{s-a}, r_2=\frac{∆}{s-b}, r_3=\frac{∆}{s-c}$

Then, $ab-r_1{r}_2=ab-\frac{{∆}^2}{\left(s-a\right)\left(s-b\right)}$

$=ab-\frac{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}{\left(s-a\right)\left(s-b\right)}$

$=ab-s\left(s-c\right)$

$=ab-\left(\frac{a+b+c}{2}\right)\left(\frac{a+b-c}{2}\right)$

$=\frac{4ab-{\left(a+b\right)}^2+c^2}{4}$

$=\frac{c^2-{\left(a-b\right)}^2}{4}$

$=\frac{c+a-b}{2}\times \frac{c-a+b}{2}$

$=\left(s-b\right)\left(s-a\right)$

Now, $\frac{ab-r_1{r}_2}{r_3}=\frac{\left(s-b\right)\left(s-a\right)}{{\displaystyle \frac{∆}{\left(s-c\right)}}}$

$=\frac{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}{s∆}$

$=\frac{∆}{s}$

$=r$