On the sides $ BC,CA,AB $ of a triangle $ABC$ semicircles $c_1,c_2,c_3$ are described externally.
If $t_1,t_2,t_3$ are the lengths of common tangents of $c_2,c_3;\;c_3,c_1$ and $c_1,c_2$ then $t_1t_2t_3$ in terms of semiperimeter and area of triangle is?
If we are able to find $t_1$ in terms of sides then $t_2 ,t_3$ will also be found similarly. We can then use relation $\Delta=\sqrt{s(s-a)(s-b)(s-c)}$.
But how to find $t_1$?
From the hints,consider evaluating $t_1$,I found the right angled triangle with sides $t_1, \frac{(b-c)}{2}$ and hypotenuse $\frac{a}{2}$. so by pythagoras' theorem
$(t_1)^2=(\frac{a}{2})^2-(\frac{(b-c)}{2})^2$ . so $t_1=\frac{(a-b+c)(a+b-c)}{4}=(s-b)(s-c)$.
Similarly $(t_2)^2=(s-c)(s-a)$,
$(t_3)^2=(s-a)(s-b)$
so $t_1t_2t_3=(s-a)(s-b)(s-c)$ $\Longleftrightarrow$ $ \frac {\Delta^2} {s}$