QUESTION IMAGE
Question
given: \\( \overrightarrow{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} \\)
Step1: Analyze \(\angle CED\)
Since \(EC\perp ED\) (right - angle symbol at \(E\) between \(EC\) and \(ED\)), by the definition of a right - angle (\(90^{\circ}\) angle), \(\angle CED = 90^{\circ}\), so \(\angle CED\) is a right angle.
Step2: Analyze \(\angle CEA\)
\(\angle CEA\) is not a right angle. A right angle is \(90^{\circ}\), and \(\angle CEA=\angle CEB+\angle BEA\).
Step3: Analyze \(m\angle CEA=\frac{1}{2}(m\angle CEB)\)
Since \(\overrightarrow{EB}\) bisects \(\angle AEC\), by the angle - bisector definition \(m\angle CEB = m\angle BEA\) and \(m\angle CEA=m\angle CEB + m\angle BEA=2m\angle CEB\), so \(m\angle CEA=\frac{1}{2}(m\angle CEB)\) is false.
Step4: Analyze \(m\angle CEB=m\angle BEA\)
Because \(\overrightarrow{EB}\) bisects \(\angle AEC\), by the angle - bisector definition (a ray that divides an angle into two congruent adjacent angles), \(m\angle CEB = m\angle BEA\).
Step5: Analyze \(m\angle DEB\)
Since \(\angle CED = 90^{\circ}\) and \(\angle CEB=\angle BEA\), and \(\angle AEC+\angle CED = 180^{\circ}\) (linear pair), \(\angle AEC = 90^{\circ}\), then \(\angle CEB=\angle BEA = 45^{\circ}\). Using the linear - pair relationship \(\angle DEB=\angle DEC+\angle CEB\), and since \(\angle DEC = 90^{\circ}\) and \(\angle CEB = 45^{\circ}\), \(m\angle DEB=90^{\circ}+45^{\circ}=135^{\circ}\).
Step6: Analyze \(m\angle AEB\)
Since \(\angle AEC = 90^{\circ}\) and \(\overrightarrow{EB}\) bisects \(\angle AEC\), \(m\angle AEB=\frac{1}{2}m\angle AEC = 45^{\circ}
eq35^{\circ}\)
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\(\angle CED\) is a right angle, \(m\angle CEB = m\angle BEA\), \(m\angle DEB = 135^{\circ}\)