Question

in the diagram below, \\( \overline { be } \cong \overline { ec } \\), \\( m \angle aeb = 20 ^ { \circ } \\) and \\( m \angle ecd = 117 ^ { \circ } \\). find \\( m \angle a \\).

Explanation:

Step1: Find ∠ECB

Since ∠ECD and ∠ECB are supplementary (they form a linear pair), we know that \( m\angle ECD + m\angle ECB = 180^\circ \). Given \( m\angle ECD = 117^\circ \), we can solve for \( m\angle ECB \):
\( m\angle ECB = 180^\circ - 117^\circ = 63^\circ \)

Step2: Analyze triangle EBC

Given \( \overline{BE} \cong \overline{EC} \), triangle EBC is isosceles with \( \angle EBC = \angle ECB \)? Wait, no, \( BE = EC \), so the base angles are \( \angle EBC \) and \( \angle ECB \)? Wait, no, in triangle EBC, sides \( BE = EC \), so the angles opposite them are \( \angle ECB \) and \( \angle EBC \) respectively. Wait, actually, \( BE = EC \), so \( \angle EBC = \angle ECB \)? Wait, no, side \( BE \) is opposite \( \angle ECB \), and side \( EC \) is opposite \( \angle EBC \). So if \( BE = EC \), then \( \angle ECB = \angle EBC \). Wait, but we found \( \angle ECB = 63^\circ \), so \( \angle EBC = 63^\circ \)? Wait, no, maybe I mixed up. Wait, actually, \( BE = EC \), so triangle EBC is isosceles with \( BE = EC \), so the base is \( BC \), and the equal sides are \( BE \) and \( EC \). Therefore, the base angles are \( \angle EBC \) and \( \angle ECB \)? Wait, no, the angles opposite the equal sides: \( BE \) is opposite \( \angle ECB \), \( EC \) is opposite \( \angle EBC \). So if \( BE = EC \), then \( \angle ECB = \angle EBC \). So \( \angle EBC = \angle ECB = 63^\circ \)? Wait, but let's check the sum of angles in triangle EBC: \( \angle BEC + \angle EBC + \angle ECB = 180^\circ \). Wait, but we need to find \( \angle BEC \). Wait, maybe another approach.

Wait, we know \( m\angle AEB = 20^\circ \), and we need to find \( m\angle A \). Let's consider triangle AEB or triangle AEC. Wait, maybe first find \( \angle BEC \). Wait, since \( \angle ECB = 63^\circ \) (from step 1, since \( \angle ECD = 117^\circ \), linear pair), and \( BE = EC \), so triangle EBC is isosceles with \( BE = EC \), so \( \angle EBC = \angle ECB = 63^\circ \)? Wait, no, that can't be, because then \( \angle BEC = 180^\circ - 63^\circ - 63^\circ = 54^\circ \). Wait, but \( \angle AEB = 20^\circ \), so \( \angle AEC = \angle AEB + \angle BEC = 20^\circ + 54^\circ = 74^\circ \)? No, maybe not. Wait, maybe I made a mistake. Wait, let's re-examine.

Wait, \( \angle ECD = 117^\circ \), so \( \angle ECB = 180^\circ - 117^\circ = 63^\circ \) (linear pair). Now, since \( BE = EC \), triangle EBC is isosceles with \( BE = EC \), so \( \angle EBC = \angle ECB = 63^\circ \)? Wait, no, \( BE = EC \), so the angles opposite are \( \angle ECB \) (opposite BE) and \( \angle EBC \) (opposite EC). So yes, \( \angle ECB = \angle EBC = 63^\circ \). Then, in triangle EBC, \( \angle BEC = 180^\circ - 63^\circ - 63^\circ = 54^\circ \). Now, \( \angle AEB = 20^\circ \), so \( \angle AEC = \angle AEB + \angle BEC = 20^\circ + 54^\circ = 74^\circ \)? Wait, no, maybe \( \angle AEB \) is part of \( \angle AEC \)? Wait, maybe the diagram shows that E is a point, and AEB is 20 degrees, and EBC is related. Wait, maybe another approach: since \( BE = EC \), triangle EBC is isosceles, so \( \angle EBC = \angle ECB = 63^\circ \). Then, \( \angle ABE = 180^\circ - \angle EBC = 180^\circ - 63^\circ = 117^\circ \)? No, that doesn't make sense. Wait, maybe I messed up the diagram. Let's assume that points A, B, C are colinear? Wait, the diagram shows D, C, B, A colinear? So DCB A is a straight line? So D---C---B---A is a straight line. So \( \angle ECD = 117^\circ \), so \( \angle ECB = 180 - 117 = 63^\circ \), as before. Now, \( BE = EC \), so triangle EBC is isosceles with \( BE = EC \), so \( \angle EBC = \angle ECB = 63^\circ \). Then, in triangle AEB, we have \( \angle AEB = 20^\circ \), and we need to find \( \angle A \). Wait, maybe triangle AEB and triangle AEC have some relation. Wait, maybe \( AE \) is a common side, and \( BE = EC \), and maybe \( \angle AEB = 20^\circ \), so let's find \( \angle A \). Wait, in triangle AEB, the sum of angles is 180. Wait, but we need to find \( \angle A \), so we need \( \angle ABE \). Wait, \( \angle ABE = 180^\circ - \angle EBC = 180^\circ - 63^\circ = 117^\circ \)? No, that can't be, because then \( \angle A + 20^\circ + 117^\circ = 180^\circ \), so \( \angle A = 43^\circ \)? Wait, that might be it. Wait, let's check:

1. \( \angle ECB = 180^\circ - 117^\circ = 63^\circ \) (linear pair)
2. \( BE = EC \), so \( \angle EBC = \angle ECB = 63^\circ \) (isosceles triangle)
3. \( \angle ABE = 180^\circ - \angle EBC = 180^\circ - 63^\circ = 117^\circ \) (linear pair, since A, B, C are colinear)
4. In triangle AEB, \( \angle A + \angle AEB + \angle ABE = 180^\circ \)
5. So \( \angle A = 180^\circ - 20^\circ - 117^\circ = 43^\circ \)

Wait, that seems to work. Let's verify:

- \( \angle ECB = 63^\circ \), \( BE = EC \), so \( \angle EBC = 63^\circ \)
- \( \angle ABE = 180 - 63 = 117^\circ \) (since A, B, C are colinear, \( \angle ABE \) and \( \angle EBC \) are supplementary)
- In triangle AEB: \( \angle A + 20 + 117 = 180 \)
- \( \angle A = 180 - 137 = 43^\circ \)

Yes, that makes sense.

Answer:

\( 43^\circ \)