To solve for \( m\angle C \), we will use the information provided about the segments and the angle.
Given that:
- \( \overline{AB} \cong \overline{BD} \cong \overline{BC} \)
- \( m\angle A = 26^\circ \)
From the congruence of the segments, we can determine that \( AB \), \( BD \), and \( BC \) are all equal in length. This implies that triangle \( ABD \) and triangle \( BCD \) may have properties that can help us find angle \( C \).
Since \( \overline{AB} \cong \overline{BD} \cong \overline{BC} \), it seems that triangles \( ABD \) and \( BCD \) could be both isosceles triangles. In particular, triangle \( ABD \) is isosceles with \( AB = BD \). Hence, if \( m\angle A = 26^\circ \), then angles \( \angle ABD \) and \( \angle ADB \) will be equal.
Let \( m \angle ABD = m \angle ADB = x \). Using the triangle angle sum property for triangle \( ABD \):
\[ m\angle A + m\angle ABD + m\angle ADB = 180^\circ \] \[ 26^\circ + x + x = 180^\circ \] \[ 26^\circ + 2x = 180^\circ \] \[ 2x = 180^\circ - 26^\circ \] \[ 2x = 154^\circ \] \[ x = 77^\circ \]
Thus, \( m\angle ABD = m\angle ADB = 77^\circ \).
Now, consider triangle \( BCD \). Since \( BC \cong BD \), triangle \( BCD \) is also isosceles, and so \( m\angle BDC = m\angle BCD \).
Let \( m\angle BDC = m\angle BCD = y \):
Again apply the triangle angle sum property for triangle \( BCD \):
\[ m\angle B + m\angle BDC + m\angle BCD = 180^\circ \]
Since \( m\angle B = m\angle ABD = 77^\circ \):
\[ 77^\circ + y + y = 180^\circ \] \[ 77^\circ + 2y = 180^\circ \] \[ 2y = 180^\circ - 77^\circ \] \[ 2y = 103^\circ \] \[ y = 51.5^\circ \]
Thus, \( m\angle BDC = m\angle BCD = 51.5^\circ \).
Since \( m\angle C = m\angle BCD \):
\[ m\angle C = 51.5^\circ \]
Thus, the final result is:
\[ \boxed{51.5^\circ} \]