Question

given: \\( \overleftrightarrow{ad} \\),
\\( \overrightarrow{eb} \\) bisects \\( \angle aec \\).
what statements are true regarding the given statement
and diagram?
\\( \square \angle ced \\) is a right angle.
\\( \square \angle cea \\) is a right angle.
\\( \square m \angle cea = \frac { 1 } { 2 } ( m \angle ceb ) \\)
\\( \square m \angle ceb = m \angle bea \\)
\\( \square m \angle deb = 135 ^ { \circ } \\)
\\( \square m \angle aeb = 35 ^ { \circ } \\)

Explanation:

Step1: Analyze ∠CED

Since \( \overrightarrow{AD} \) is a straight line and \( EC\perp ED \) (right - angle symbol at \( E \)), \( \angle CED = 90^{\circ}\), so \( \angle CED \) is a right - angle.

Step2: Analyze ∠CEA

From the diagram, \( \angle CEA\) is not a right - angle.

Step3: Analyze \(m\angle CEA=\frac{1}{2}(m\angle CEB)\)

Since \( \overrightarrow{EB}\) bisects \( \angle AEC\), \(m\angle CEB=m\angle BEA\) and \(m\angle CEA = 2m\angle CEB\), so \(m\angle CEA=\frac{1}{2}(m\angle CEB)\) is false.

Step4: Analyze \(m\angle CEB = m\angle BEA\)

By the definition of an angle bisector (\(\overrightarrow{EB}\) bisects \( \angle AEC\)), \(m\angle CEB = m\angle BEA\).

Step5: Analyze \(m\angle DEB\)

Since \(m\angle CEB=m\angle BEA = 45^{\circ}\) (because \( \angle CEA=90^{\circ}\) and \( \overrightarrow{EB}\) bisects \( \angle AEC\)), and \(m\angle DEB=m\angle DEC + m\angle CEB\), \(m\angle DEB=90^{\circ}+ 45^{\circ}=135^{\circ}\)

Step6: Analyze \(m\angle AEB\)

Since \(m\angle AEB = 45^{\circ}
eq35^{\circ}\)

Answer:

\(\angle CED\) is a right angle, \(m\angle CEB = m\angle BEA\), \(m\angle DEB = 135^{\circ}\)