$A$ and $B$ are on a circle of radius $20$ centered at $C$, and $\angle ACB = 60^\circ$. $D$ is chosen so that $D$ is also on the circle, $\angle ACD = 160^\circ$, and $\angle DCB = 100^\circ$. Let $E$ be the intersection of lines $AC$ and $BD$. What is $DE$?
1. Given that $A$ and $B$ are on a circle of radius $20$ centered at $C$, and $\angle ACB = 60^\circ$. We need to find the length of $DE$ where $D$ is also on the circle, $\angle ACD = 160^\circ$, and $\angle DCB = 100^\circ$. Let $E$ be the intersection of lines $AC$ and $BD$.

2. First, we note that $\angle ACB = 60^\circ$ implies that the arc $AB$ subtends an angle of $60^\circ$ at the center $C$. Therefore, the measure of the arc $AB$ is $60^\circ$.

3. Next, we consider the angles $\angle ACD = 160^\circ$ and $\angle DCB = 100^\circ$. Since $D$ is on the circle, $\angle ACD + \angle DCB = 160^\circ + 100^\circ = 260^\circ$. This means that the arc $AD$ subtends an angle of $160^\circ$ at the center $C$.

4. Let $AC$ intersect the circle again at $F$. Then, $\angle ACF = 180^\circ$ because $F$ is the diametrically opposite point of $A$ on the circle.

5. We need to find $\angle BEA$. Using the fact that $\angle BEA$ is the angle between the chords $AB$ and $DF$, we can use the formula for the angle between two chords:
   \[
   \angle BEA = \frac{\widehat{AB} - \widehat{DF}}{2}
   \]
   Here, $\widehat{AB} = 60^\circ$ and $\widehat{DF} = 180^\circ - 160^\circ = 20^\circ$.

6. Substituting these values, we get:
   \[
   \angle BEA = \frac{60^\circ - 20^\circ}{2} = \frac{40^\circ}{2} = 20^\circ
   \]

7. Now, consider $\angle DCF$. Since $\angle ACD = 160^\circ$, we have:
   \[
   \angle DCF = 180^\circ - 160^\circ = 20^\circ
   \]

8. Since $\angle BEA = 20^\circ$ and $\angle DCF = 20^\circ$, we observe that $\triangle DEC$ is isosceles with $DE = DC$.

9. Given that $DC$ is a radius of the circle, we have $DC = 20$.

10. Therefore, $DE = 20$.

The final answer is $\boxed{20}$.