To find \( m\angle B \) given that \( AB \cong BC \) and \( m\angle BCD = 115^\circ \), we can start by recognizing that since \( AB \cong BC \), triangle \( ABC \) is isosceles with \( AB = BC \).
In an isosceles triangle, the angles opposite the equal sides are also equal. Thus, we can say:
\[ m\angle A = m\angle B \]
Now, let’s denote \( m\angle A = m\angle B = x \). The angles in triangle \( ABC \) add up to \( 180^\circ \):
\[ x + x + m\angle ABC = 180^\circ \]
Since angle \( BCD \) is the exterior angle of triangle \( ABC \), we can use the exterior angle theorem, which states that an exterior angle is equal to the sum of the two opposing interior angles:
\[ m\angle BCD = m\angle A + m\angle B \]
This can be written as:
\[ 115^\circ = x + x = 2x \]
Now, we can solve for \( x \):
\[ 2x = 115^\circ \ x = \frac{115^\circ}{2} \ x = 57.5^\circ \]
Since \( x = m\angle B \), we have:
\[ m\angle B = 57.5^\circ \]
Thus, the measure of angle \( B \) is:
\[ \boxed{57.5^\circ} \]
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