Given that \( \overline{BC} \cong \overline{EC} \), we can infer that triangle \( BEC \) is isosceles, with \( BC = EC \).
Since \( \overline{BC} \cong \overline{EC} \), the angles opposite these sides will be equal. Thus, we have:
\[ m \angle BEC = m \angle EBC \]
Let's denote these angles as \( x \). Since the angles of triangle \( BEC \) sum up to \( 180^\circ \), we can write:
\[ m \angle BEC + m \angle EBC + m \angle CBE = 180^\circ \]
This becomes:
\[ x + x + m \angle CBE = 180^\circ \]
\[ 2x + m \angle CBE = 180^\circ \]
Next, we also need to consider triangle \( DCF \). From the information given, we know:
\[ m \angle DCF = 44^\circ \] Let’s denote \( m \angle CDF = y \).
Using the triangle sum property for triangle \( DCF \):
\[ m \angle DCF + m \angle CDF + m \angle FDC = 180^\circ \]
This is:
\[ 44^\circ + y + m \angle FDC = 180^\circ \]
Solving for \( m \angle FDC \):
\[ m \angle FDC = 180^\circ - 44^\circ - y = 136^\circ - y \]
To find \( m \angle ABE \), we consider the relationship between angles \( m \angle CBE \) and \( m \angle FDC \). Since \( A, B, C, D, E, F \) suggest a certain cyclical order in triangle and angle placement:
- Angles \( \angle CBE \) and \( \angle DCF \) are supplementary:
\[ m \angle CBE + m \angle DCF = 180^\circ \]
Thus:
\[ m \angle CBE + 44^\circ = 180^\circ \]
\[ m \angle CBE = 136^\circ \]
Substituting back into our equation for \( m \angle ABE \):
\[ 2x + 136^\circ = 180^\circ \]
Now, solving for \( x \):
\[ 2x = 180^\circ - 136^\circ = 44^\circ \] \[ x = 22^\circ \]
Thus, \( m \angle ABE = m \angle CBE \), so:
\[ m \angle ABE = 22^\circ \]
To conclude, we find the measure of angle \( ABE \):
\[ \boxed{22^\circ} \]