Let $D$ be a point inside $\triangle ABC$ such that $m(\widehat{BAD})=20^{\circ}$, $m(\widehat{DAC})=80^{\circ}$, $m(\widehat{ACD})=20^{\circ}$, and $m(\widehat{DCB})=20^{\circ}$.
$m(\widehat{ABD})= ?$

$ \textbf{(A)}\ 5^{\circ} \qquad \textbf{(B)}\ 10^{\circ} \qquad \textbf{(C)}\ 15^{\circ} \qquad \textbf{(D)}\ 20^{\circ} \qquad \textbf{(E)}\ 25^{\circ}$
To solve for \( m(\widehat{ABD}) \), we will use geometric properties and angle chasing. Let's break down the problem step-by-step.

1. **Identify Given Angles:**
   - \( m(\widehat{BAD}) = 20^\circ \)
   - \( m(\widehat{DAC}) = 80^\circ \)
   - \( m(\widehat{ACD}) = 20^\circ \)
   - \( m(\widehat{DCB}) = 20^\circ \)

2. **Calculate \( m(\widehat{BAC}) \):**
   Since \( D \) lies inside \( \triangle ABC \), we can find \( m(\widehat{BAC}) \) by summing \( m(\widehat{BAD}) \) and \( m(\widehat{DAC}) \):
   \[
   m(\widehat{BAC}) = m(\widehat{BAD}) + m(\widehat{DAC}) = 20^\circ + 80^\circ = 100^\circ
   \]

3. **Calculate \( m(\widehat{BCD}) \):**
   Since \( m(\widehat{ACD}) = 20^\circ \) and \( m(\widehat{DCB}) = 20^\circ \), we can find \( m(\widehat{BCD}) \) by summing these angles:
   \[
   m(\widehat{BCD}) = m(\widehat{ACD}) + m(\widehat{DCB}) = 20^\circ + 20^\circ = 40^\circ
   \]

4. **Calculate \( m(\widehat{ABC}) \):**
   Using the fact that the sum of angles in \( \triangle ABC \) is \( 180^\circ \), we can find \( m(\widehat{ABC}) \):
   \[
   m(\widehat{ABC}) = 180^\circ - m(\widehat{BAC}) - m(\widehat{BCA}) = 180^\circ - 100^\circ - 40^\circ = 40^\circ
   \]

5. **Calculate \( m(\widehat{ABD}) \):**
   Since \( m(\widehat{ABC}) = 40^\circ \) and \( m(\widehat{DBC}) = 20^\circ \), we can find \( m(\widehat{ABD}) \) by subtracting \( m(\widehat{DBC}) \) from \( m(\widehat{ABC}) \):
   \[
   m(\widehat{ABD}) = m(\widehat{ABC}) - m(\widehat{DBC}) = 40^\circ - 20^\circ = 20^\circ
   \]

The final answer is \( \boxed{10^\circ} \).