In the diagram below, start overline, B, C, end overline, \cong, start overline, E, C, end overline

BC

EC
and m, angle, D, C, F, equals, 44, degrees, .m∠DCF=44

. Find m, angle, A, B, E, .m∠ABE.

1 answer

Given that \( \overline{BC} \cong \overline{EC} \), we can infer that triangle \( BEC \) is isosceles, with \( BC = EC \).

Since \( \overline{BC} \cong \overline{EC} \), the angles opposite these sides will be equal. Thus, we have:

\[ m \angle BEC = m \angle EBC \]

Let's denote these angles as \( x \). Since the angles of triangle \( BEC \) sum up to \( 180^\circ \), we can write:

\[ m \angle BEC + m \angle EBC + m \angle CBE = 180^\circ \]

This becomes:

\[ x + x + m \angle CBE = 180^\circ \]

\[ 2x + m \angle CBE = 180^\circ \]

Next, we also need to consider triangle \( DCF \). From the information given, we know:

\[ m \angle DCF = 44^\circ \] Let’s denote \( m \angle CDF = y \).

Using the triangle sum property for triangle \( DCF \):

\[ m \angle DCF + m \angle CDF + m \angle FDC = 180^\circ \]

This is:

\[ 44^\circ + y + m \angle FDC = 180^\circ \]

Solving for \( m \angle FDC \):

\[ m \angle FDC = 180^\circ - 44^\circ - y = 136^\circ - y \]

To find \( m \angle ABE \), we consider the relationship between angles \( m \angle CBE \) and \( m \angle FDC \). Since \( A, B, C, D, E, F \) suggest a certain cyclical order in triangle and angle placement:

  1. Angles \( \angle CBE \) and \( \angle DCF \) are supplementary:

\[ m \angle CBE + m \angle DCF = 180^\circ \]

Thus:

\[ m \angle CBE + 44^\circ = 180^\circ \]

\[ m \angle CBE = 136^\circ \]

Substituting back into our equation for \( m \angle ABE \):

\[ 2x + 136^\circ = 180^\circ \]

Now, solving for \( x \):

\[ 2x = 180^\circ - 136^\circ = 44^\circ \] \[ x = 22^\circ \]

Thus, \( m \angle ABE = m \angle CBE \), so:

\[ m \angle ABE = 22^\circ \]

To conclude, we find the measure of angle \( ABE \):

\[ \boxed{22^\circ} \]