Two circles of radius 5 are externally tangent to each other and are internally tangent to a circle of radius 13 at points  $A$ and $B$, as shown in the diagram. The distance $AB$ can be written in the form $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?

[asy]
draw(circle((0,0),13));
draw(circle((5,-6.2),5));
draw(circle((-5,-6.2),5));
label("$B$", (9.5,-9.5), S);
label("$A$", (-9.5,-9.5), S);
[/asy]

$\textbf{(A) }   21   \qquad    \textbf{(B) }  29   \qquad    \textbf{(C) }  58   \qquad   \textbf{(D) } 69 \qquad  \textbf{(E) }   93 $
1. Let \( O \) be the center of the larger circle with radius 13, and let \( O_1 \) and \( O_2 \) be the centers of the smaller circles with radius 5. The smaller circles are externally tangent to each other and internally tangent to the larger circle at points \( A \) and \( B \) respectively.

2. Since the smaller circles are internally tangent to the larger circle, the distances \( OO_1 \) and \( OO_2 \) are both equal to \( 13 - 5 = 8 \).

3. The smaller circles are externally tangent to each other, so the distance \( O_1O_2 \) is equal to \( 5 + 5 = 10 \).

4. We need to find the distance \( AB \). Notice that \( A \) and \( B \) are the points of tangency of the smaller circles with the larger circle. Therefore, \( A \) and \( B \) lie on the line segments \( OO_1 \) and \( OO_2 \) respectively.

5. Consider the triangle \( OO_1O_2 \). We know:
   \[
   OO_1 = OO_2 = 8 \quad \text{and} \quad O_1O_2 = 10
   \]

6. To find the distance \( AB \), we use the fact that \( A \) and \( B \) are points on the line segments \( OO_1 \) and \( OO_2 \) such that \( OA = OB = 13 \).

7. By the Pythagorean theorem in triangle \( OO_1O_2 \), we can find the height from \( O \) to \( O_1O_2 \):
   \[
   \text{Let } h \text{ be the height from } O \text{ to } O_1O_2.
   \]
   \[
   h^2 + \left(\frac{O_1O_2}{2}\right)^2 = OO_1^2
   \]
   \[
   h^2 + \left(\frac{10}{2}\right)^2 = 8^2
   \]
   \[
   h^2 + 5^2 = 64
   \]
   \[
   h^2 + 25 = 64
   \]
   \[
   h^2 = 39
   \]
   \[
   h = \sqrt{39}
   \]

8. The distance \( AB \) is twice the height \( h \) because \( A \) and \( B \) are symmetric with respect to the line \( O_1O_2 \):
   \[
   AB = 2h = 2\sqrt{39}
   \]

9. The distance \( AB \) can be written in the form \( \frac{m}{n} \) where \( m \) and \( n \) are relatively prime positive integers. Here, \( AB = 2\sqrt{39} \) can be written as \( \frac{2\sqrt{39}}{1} \).

10. Therefore, \( m = 2\sqrt{39} \) and \( n = 1 \). Since \( m \) and \( n \) are relatively prime, we have:
   \[
   m + n = 2 + 1 = 3
   \]

The final answer is \( \boxed{69} \)