We have

$r{r}_2=r_1{r}_3$

We know the formula for in-radius and ex-radius

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

Applying

$r{r}_2=r_1{r}_3$

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

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

$\Rightarrow s^2{\left(s-b\right)}^2=s\left(s-b\right)\left(s-c\right)\left(s-a\right)$

$\Rightarrow s^2{\left(s-b\right)}^2={∆}^2$

$\Rightarrow {\left(\frac{∆}{s\left(s-a\right)}\right)}^2=1 \Rightarrow {\tan }^2\left(\frac{B}{2}\right)=1 \Rightarrow \frac{B}{2}=\frac{\mathrm{\pi }}{4}, \frac{5\mathrm{\pi }}{4} \Rightarrow B=\frac{\mathrm{\pi }}{2}, \frac{5\mathrm{\pi }}{2}$

$\Rightarrow \cos 2B=\cos \mathrm{\pi } \mathrm{or} \cos 5\mathrm{\pi }=-1$