Let the $A$-exradii be $r_A,$ and define $r_B,r_C$ in the same manner. Also, let $r$ be the length of the inradius of $ABC.$ Thus, \begin{align*} \frac{s}{s-a}\cdot r&=r_A,\tag{1}\\ \frac{s}{s-b}\cdot r&=r_B,\text{ and }\tag{2}\\ \frac{s}{s-c}\cdot r&=r_C.\tag{3} \end{align*}Dividing equations (1) and (2) and equations (1) and (3), we find that \begin{align*} \frac{s-b}{s-a}&=\frac{r_B}{r_A}\text{ and }\tag{4}\\ \frac{s-c}{s-a}&=\frac{r_C}{r_A}.\tag{5} \end{align*}Adding equations (4) and (5), we find that \begin{align*} \frac{a}{s-a}=\frac{r_B+r_C}{r_A}&\Leftrightarrow\\ \frac{s}{s-a}=\frac{r_B+r_C+r_A}{r_A}&\Leftrightarrow\\ s-a=s\cdot \frac{r_A}{r_A+r_B+r_C}&\Leftrightarrow\\ a=s\cdot\frac{r_B+r_C}{r_A+r_B+r_C}&. \end{align*}Therefore, we have know know that $a:b:c=r_B+r_C:r_A+r_C:r_A+r_B.$ Now, we can construct the triangle using standard techniques. (Fix $BC,$ draw circles of radius $AB,AC$ centered at $B,C,$ respectively. Their intersection (it doesn't matter which) is point $A$. Thus, we have all three vertices, which means that we have constructed the triangle).