Question

You have an equilateral triangle BCD and an isosceles triangle
BAD where BA = BD and they share a common side BD.
m A = 61 Determine the measure of ABC

Answer 1

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:

\[
\boxed{119.5^\circ}
\]