A circle of radius $4$ is inscribed in a triangle $ABC$. We call $D$ the touchpoint between the circle and side BC. Let $CD =8$, $DB= 10$. What is the length of the sides $AB$ and $AC$?
1. Let the sides of the triangle be denoted as follows: \( AB = c \), \( BC = a \), and \( AC = b \). The semi-perimeter \( s \) of the triangle is given by:
   \[
   s = \frac{a + b + c}{2}
   \]
   Given that the circle is inscribed in the triangle, the distances from the touchpoint \( D \) to the vertices \( B \) and \( C \) are \( DB = 10 \) and \( CD = 8 \), respectively. Therefore, \( a = BC = DB + CD = 10 + 8 = 18 \).

2. The distances from the touchpoints to the vertices are related to the semi-perimeter \( s \) as follows:
   \[
   s - b = 10 \quad \text{and} \quad s - c = 8
   \]
   Adding these two equations, we get:
   \[
   (s - b) + (s - c) = 10 + 8 \implies 2s - (b + c) = 18
   \]
   Since \( s = \frac{a + b + c}{2} \), we can substitute \( a = 18 \) into the equation:
   \[
   2s - (b + c) = 18 \implies 2s - (b + c) = 18
   \]
   Subtracting the above equation from \( 2s = a + b + c \), we get:
   \[
   a = 18
   \]
   Therefore, we have:
   \[
   b + c = 2s - 18
   \]

3. Subtracting the two equations \( s - b = 10 \) and \( s - c = 8 \), we get:
   \[
   (s - b) - (s - c) = 10 - 8 \implies c - b = 2
   \]

4. Now, we use Heron's formula to find the area of the triangle. The area \( \Delta \) of the triangle can also be expressed in terms of the inradius \( r \) and the semi-perimeter \( s \):
   \[
   \Delta = r \cdot s
   \]
   Given that the radius \( r = 4 \), we have:
   \[
   \Delta = 4s
   \]

5. Using Heron's formula, the area \( \Delta \) is also given by:
   \[
   \Delta = \sqrt{s(s - a)(s - b)(s - c)}
   \]
   Equating the two expressions for the area, we get:
   \[
   4s = \sqrt{s(s - a)(s - b)(s - c)}
   \]
   Squaring both sides, we obtain:
   \[
   (4s)^2 = s(s - a)(s - b)(s - c) \implies 16s^2 = s(s - a)(s - b)(s - c)
   \]
   Dividing both sides by \( s \), we get:
   \[
   16s = (s - a)(s - b)(s - c)
   \]

6. Substituting \( s - a = s - 18 \), \( s - b = 10 \), and \( s - c = 8 \), we have:
   \[
   16s = (s - 18) \cdot 10 \cdot 8
   \]
   Simplifying, we get:
   \[
   16s = 80(s - 18)
   \]
   Solving for \( s \), we get:
   \[
   16s = 80s - 1440 \implies 64s = 1440 \implies s = \frac{1440}{64} = 22.5
   \]

7. Now, we have \( s = 22.5 \). Using the equations \( s - b = 10 \) and \( s - c = 8 \), we find:
   \[
   b = s - 10 = 22.5 - 10 = 12.5
   \]
   \[
   c = s - 8 = 22.5 - 8 = 14.5
   \]

8. Therefore, the lengths of the sides \( AB \) and \( AC \) are:
   \[
   AB = c = 14.5 \quad \text{and} \quad AC = b = 12.5
   \]

The final answer is \( AB = \boxed{14.5} \) and \( AC = \boxed{12.5} \).