To find \( m\angle A \), we start by analyzing the given information. We know:
- \( \overline{BE} \cong \overline{EC} \) means \( BE = EC \).
- \( m\angle AEB = 20^\circ \).
- \( m\angle ECD = 117^\circ \).
Since \( BE = EC \), triangle \( ABE \) is isosceles with \( BE = EC \). In triangle \( ABE \), we have the following angles:
- \( m\angle AEB = 20^\circ \)
- Since \( BE = EC \), the angles opposite these sides are equal, which means \( m\angle ABE = m\angle A \).
Now let's denote \( m\angle A = x \). The angle sum in triangle \( ABE \) gives us:
\[ m\angle A + m\angle AEB + m\angle ABE = 180^\circ \]
Substituting the known values, we have:
\[ x + 20^\circ + x = 180^\circ \]
This simplifies to:
\[ 2x + 20^\circ = 180^\circ \]
Now, subtract \( 20^\circ \) from both sides:
\[ 2x = 160^\circ \]
Dividing both sides by 2 gives us:
\[ x = 80^\circ \]
So, \( m\angle A = 80^\circ \).
Next, we also need to check the relationship between angles \( A, B, C, D \). Since \( EC \) is a continuation from \( E \) through \( C \) to \( D \), and we have:
\[ m\angle ECD = 117^\circ, \] we notice that the angles \( AEB \) and \( ECD \) should also sum up to a straight line:
\[ m\angle AEB + m\angle ECD + m\angle A = 180^\circ. \]
Substituting the known values \( 20^\circ \) and the calculated \( m\angle A = 80^\circ \):
\[ 20^\circ + 117^\circ + 80^\circ = 180^\circ, \]
which verifies that our calculations are consistent.
Thus, the final answer is:
\[ \boxed{80^\circ} \]