To find \( m \angle B \), we can use the given information that \( \overline{AB} \cong \overline{BC} \) and that \( m \angle BCD = 115^\circ \).
Since \( \overline{AB} \cong \overline{BC} \), we know that triangle \( ABC \) is isosceles with \( AB = BC \). This means that the two angles opposite the equal sides (i.e., angles \( A \) and \( C \)) must be equal. Let's denote the measure of angle \( A \) as \( m \angle A \) and the measure of angle \( C \) as \( m \angle C \).
According to the triangle angle sum theorem, the sum of the angles in triangle \( ABC \) is equal to \( 180^\circ \):
\[ m \angle A + m \angle B + m \angle C = 180^\circ \]
Since \( m \angle A = m \angle C \) (as the triangle is isosceles), we can denote both angles as \( x \). Therefore, we have:
\[ x + m \angle B + x = 180^\circ \] \[ 2x + m \angle B = 180^\circ \]
We also need to consider the angle \( \angle BCD \). The exterior angle theorem tells us that the measure of an exterior angle (in this case, \( m \angle BCD \)) is equal to the sum of the measures of the two non-adjacent interior angles (i.e., \( \angle A \) and \( \angle B \)) of triangle \( ABC \).
Thus, we can say:
\[ m \angle BCD = m \angle A + m \angle B \]
Plugging the values we have:
\[ 115^\circ = x + m \angle B \]
We now have two equations:
- \( 2x + m \angle B = 180^\circ \)
- \( x + m \angle B = 115^\circ \)
We can solve these equations simultaneously. From the second equation, we rearrange it to find \( m \angle B \):
\[ m \angle B = 115^\circ - x \]
Now we substitute \( m \angle B \) in the first equation:
\[ 2x + (115^\circ - x) = 180^\circ \] \[ 2x + 115^\circ - x = 180^\circ \] \[ x + 115^\circ = 180^\circ \] \[ x = 180^\circ - 115^\circ = 65^\circ \]
Now that we have \( x = m \angle A = m \angle C = 65^\circ \), we substitute back into the equation for \( m \angle B \):
\[ m \angle B = 115^\circ - 65^\circ = 50^\circ \]
Thus, the measure of angle \( B \) is:
\[ \boxed{50^\circ} \]