Question

given \\( \overleftrightarrow{ad} \\).
\\( \overleftrightarrow{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 $\angle CED$

Since $EC\perp ED$ (from the right - angle symbol in the diagram), by the definition of a right angle, $\angle CED = 90^{\circ}$, so $\angle CED$ is a right angle.

Step2: Use the angle - bisector property

Given that $\overrightarrow{EB}$ bisects $\angle AEC$, by the angle - bisector definition, $m\angle CEB=m\angle BEA$.

Step3: Calculate $m\angle DEB$

We know that $m\angle CEB = m\angle BEA$ and $\angle AEC+\angle CED = 180^{\circ}$ (linear pair). Since $\angle CED = 90^{\circ}$, then $\angle AEC=90^{\circ}$. So $m\angle CEB=m\angle BEA = 45^{\circ}$. Then, using the linear - pair relationship for $\angle BEA$ and $\angle DEB$ ($\angle BEA+\angle DEB = 180^{\circ}$), we substitute $m\angle BEA = 45^{\circ}$ into the equation: $m\angle DEB=180^{\circ}-m\angle BEA$. So $m\angle DEB = 180 - 45=135^{\circ}$.

Step4: Check other options

• $\angle CEA$ is not a right angle. $\angle CED$ is a right angle.
• $m\angle CEA = m\angle CEB + m\angle BEA=2m\angle CEB$ (since $m\angle CEB = m\angle BEA$), so $m\angle CEA=\frac{1}{2}(m\angle CEB)$ is false.
• There is no information to suggest $m\angle AEB = 35^{\circ}$.

Answer:

• $\angle CED$ is a right angle
• $m\angle CEB = m\angle BEA$
• $m\angle DEB = 135^{\circ}$