To find the measure of angle \( ABC \) in the configuration described, we can start by analyzing the triangles involved.
Given:
- Triangle \( BCD \) is equilateral, so each angle in triangle \( BCD \) is \( 60^\circ \).
- Triangle \( BAD \) is isosceles with \( BA = BD \) and the measure of angle \( A \), which is \( m \angle A = 61^\circ \).
Since triangle \( BAD \) is isosceles and \( BA = BD \), the angles at \( A \) and \( B \) are related. The angles \( B \) and \( D \) are equal, so we can denote them as \( m \angle B = m \angle D = x \).
The sum of angles in triangle \( BAD \) is \( 180^\circ \): \[ m \angle A + m \angle B + m \angle D = 180^\circ \] \[ 61^\circ + x + x = 180^\circ \] \[ 61^\circ + 2x = 180^\circ \] \[ 2x = 180^\circ - 61^\circ \] \[ 2x = 119^\circ \] \[ x = \frac{119^\circ}{2} = 59.5^\circ \]
Thus, \[ m \angle B = 59.5^\circ \]
Now consider the position of \( C \). In triangle \( BCD \), since it is an equilateral triangle: \[ m \angle DBC = 60^\circ \]
Finally, angle \( ABC \) can be found by adding angles \( DAB \) and \( DBC \): \[ m \angle ABC = m \angle DAB + m \angle DBC \] Where \( m \angle DAB = m \angle B = 59.5^\circ \) and \( m \angle DBC = 60^\circ \):
\[ m \angle ABC = 59.5^\circ + 60^\circ = 119.5^\circ \]
Thus, the measure of angle \( ABC \) is:
\[ \boxed{119.5^\circ} \]