What is the value of $m+n$ if $\frac{m}{n}$ is the radius of the smallest of the three circles? 

Circles of radii 5, 5, 8 and $\frac{m}{n} $(the smallest circle) are
  mutually externally tangent to all circles, where $m$ and $n$ are
  relatively prime positive  integers. Find $m + n$.

Source: Bangladesh Math Olympiad 2017 Junior category.
I can not figure the radius of the smallest circle. Is there any formula for an internal point in a isosceles triangle which can help me to solve this math?

The triangle has a base side of length $10$ and two sides of length $13$ each. The hight this trinagle is $h = \sqrt {13^2-5^2} = 12$. This height consists of the radius of the large circle, the radius $r=\frac mn$ of the smallest circle and the hight of another isoceles triangle with the same base side and two other sides of lengths $5+r$, which is $\sqrt {(5+r)^2-5^2} = \sqrt{10r+r^2}$. Therefore, we have
$$
\begin{align}
& \quad 12 = 8 + r + \sqrt{10r+r^2}\\
\Leftrightarrow & \quad 4-r = \sqrt {10r+r^2}\\
\Leftrightarrow & \quad (4-r)^2 = 10r + r^2\\
\Leftrightarrow & \quad 16 - 8r + r^2 = 10r + r^2\\
\Leftrightarrow & \quad 18r = 16\\
\Leftrightarrow & \quad r = \frac{16}{18} = \frac89\\
\end{align}
$$
Thus $m=8$ and $n=9$, so $n+m=17$.
