Triangular Lemma!

Here is a pretty messy solution using coordinate geometry:

Position $A$ at the origin $O$ such that $AB$ is on the $x$ -axis and $AC$ is on the $y$ -axis; then $S_1$ has the equation $(x-r_1)^2+(y-r_1)^2=r_1^2$ and $S_2$ has the equation $(x-r_2)^2+(y-r_2)^2=r_2^2.$ If $R$ is the radius of the circle with the common property of $S_1$ and $S_2$ , then we have $\pm \sqrt{R^2-(x-R)^2}+R=\pm \sqrt{\dfrac{m^2+n^2}{4}-\left(x-\dfrac{m}{2}\right)^2}+\dfrac n2,$ where $AB=m$ and $AC=n$ . We can replace the $\pm$ on both sides with $+$ since it is evident that both $S_1$ and $S_2$ must intersect $S$ at a $y$ -coordinate greater than $C$ , and thus in their positive halves; not doing so would result in solutions for $R$ corresponding to the radii of the two other externally tangent circles (this ambiguity should probably be clarified in the question). Also, this equation must have one solution in $x$ since both $S_1$ and $S_2$ touch $S$ tangentially. Therefore, we can let $\Delta =0$ and simplify to obtain $\begin{aligned} -2mn^3R^2+10mn^2R^3-16mnR^4+8mR^5+2n^3R^3-9n^2R^4+12nR^5-4R^6&=0\\ \Longrightarrow R^2\left(R-\dfrac n2\right)^2\left(R+\sqrt{m^2+n^2}-m-n\right)\left(R-\sqrt{m^2+n^2}-m-n\right)&=0\\ \Longrightarrow R=0,\;\;\; \dfrac n2,\;\;\; m+n-\sqrt{m^2+n^2},\;\;\; m+n+\sqrt{m^2+n^2}. \end{aligned}$ We can discard $R=\frac n2$ due to the generality of our construction about $y=x$ , and we can discard $R=0$ (corresponding to the point $O$ ). This leaves $r_1=m+n-\sqrt{m^2+n^2}\; \text{ and }\; r_2=m+n+\sqrt{m^2+n^2}.$ Hence, the area $[\triangle ABC ]$ is given by $[\triangle ABC ]=\dfrac{mn}{2}=\dfrac{\left(m+n-\sqrt{m^2+n^2}\right)\left(m+n+\sqrt{m^2+n^2}\right)}{4}=\boxed{\dfrac{r_1r_2}{4}}.$