If in a triangle the ex-radii  r₁,r₂,r₃ are in the ratio 1:2:3, then their sides are in the ratio

$\begin{array}{l}\text{ :Let }{r}_1=\frac{\mathrm{\Delta }}{(s-a)}=k\\ r_2=\frac{\mathrm{\Delta }}{(s-b)}=2k\text{ and }{r}_3=\frac{\mathrm{\Delta }}{(s-c)}=3k\\ \therefore s-a=\frac{\mathrm{\Delta }}{k}----\left(i\right)\\ s-b=\frac{\mathrm{\Delta }}{\left(2k\right)}---\left(ii\right)\\ \text{ and }s-c=\frac{\mathrm{\Delta }}{\left(3k\right)}---\left(ii\right)\end{array}$
Adding eqns. (ii) and (iii)
$2s-b-c=\frac{\mathrm{\Delta }}{2k}+\frac{\mathrm{\Delta }}{3k}\Rightarrow a=\frac{5\mathrm{\Delta }}{6k}$
Similarly, adding (i) and (iii), (i) and (ii),
$\begin{array}{l}b=\frac{4\mathrm{\Delta }}{3k}\text{ and }c=\frac{3\mathrm{\Delta }}{2k}\\ \therefore a:b:c=\frac{5\mathrm{\Delta }}{6k}:\frac{4\mathrm{\Delta }}{3k}:\frac{3\mathrm{\Delta }}{2k}=5:8:9\end{array}$