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

BE

EC
, m, angle, A, E, B, equals, 20, degreesm∠AEB=20

and m, angle, E, C, D, equals, 117, degrees, .m∠ECD=117

. Find m, angle, A, .m∠A.

1 answer

To find \( m\angle A \), we start by analyzing the given information. We know:

  1. \( \overline{BE} \cong \overline{EC} \) means \( BE = EC \).
  2. \( m\angle AEB = 20^\circ \).
  3. \( m\angle ECD = 117^\circ \).

Since \( BE = EC \), triangle \( ABE \) is isosceles with \( BE = EC \). In triangle \( ABE \), we have the following angles:

  • \( m\angle AEB = 20^\circ \)
  • Since \( BE = EC \), the angles opposite these sides are equal, which means \( m\angle ABE = m\angle A \).

Now let's denote \( m\angle A = x \). The angle sum in triangle \( ABE \) gives us:

\[ m\angle A + m\angle AEB + m\angle ABE = 180^\circ \]

Substituting the known values, we have:

\[ x + 20^\circ + x = 180^\circ \]

This simplifies to:

\[ 2x + 20^\circ = 180^\circ \]

Now, subtract \( 20^\circ \) from both sides:

\[ 2x = 160^\circ \]

Dividing both sides by 2 gives us:

\[ x = 80^\circ \]

So, \( m\angle A = 80^\circ \).

Next, we also need to check the relationship between angles \( A, B, C, D \). Since \( EC \) is a continuation from \( E \) through \( C \) to \( D \), and we have:

\[ m\angle ECD = 117^\circ, \] we notice that the angles \( AEB \) and \( ECD \) should also sum up to a straight line:

\[ m\angle AEB + m\angle ECD + m\angle A = 180^\circ. \]

Substituting the known values \( 20^\circ \) and the calculated \( m\angle A = 80^\circ \):

\[ 20^\circ + 117^\circ + 80^\circ = 180^\circ, \]

which verifies that our calculations are consistent.

Thus, the final answer is:

\[ \boxed{80^\circ} \]