Centers of internally tangent circles lie along radii of circle $A;$ put the center of $A$ at the origin with the radius-$3$ vertex on top

By symmetry $E$’s center is on the vertical axis. Tangency to $B$ puts it at height $4 - r;$ then tangency to $C$ gives an equation in $r.$

Place the center of circle $A$ at the origin with the triangle’s vertices at $(0, 10)$ and $\left(\pm 5\sqrt{3}, -5\right).$ A circle internally tangent to $A$ at a vertex has its center on the radius to that vertex, so circle $B$ has center $(0, 7)$ and circles $C$ and $D$ have centers $\left(\mp 4\sqrt{3}, -4\right)$ (at distance $10 - 2 = 8$ from the origin).

By symmetry the center of circle $E,$ of radius $r,$ lies on the $y$-axis at $(0, y).$ External tangency to $B$ gives $7 - y = r + 3,$ so $y = 4 - r.$ External tangency to $C$ gives $$\begin{aligned} &\left(4\sqrt{3}\right)^2 \\ &\quad {}+ (4 - r + 4)^2 = (r + 2)^2, \end{aligned}$$ that is, $48 + (8 - r)^2 = (r + 2)^2,$ which simplifies to $112 - 16r = 4r + 4,$ so $r = \frac{27}{5}.$

Then $m + n = 27 + 5 = 32.$