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identifying relationships from diagrams given: $overrightarrow{eb}$ bis…

Question

identifying relationships from diagrams
given: $overrightarrow{eb}$ bisects $angle aec$.
$angle aed$ is a straight angle.
prove: $mangle aeb = 45^circ$
complete the paragraph proof.
we are given that $overrightarrow{eb}$ bisects $angle aec$. from the diagram, $angle ced$ is a right angle, which measures degrees. since the measure of a straight angle is $180^circ$, the mea must also be $90^circ$ by the a bisector cuts the angle measure in half $mangle aeb = 45^circ$.

Explanation:

Step1: Recall right angle measure

A right angle measures \( 90^\circ \), so \( \angle CED = 90^\circ \).

Step2: Analyze straight angle \( \angle AED \)

Since \( \angle AED \) is a straight angle (\( 180^\circ \)), \( \angle AEC + \angle CED = 180^\circ \). Substituting \( \angle CED = 90^\circ \), we get \( \angle AEC = 180^\circ - 90^\circ = 90^\circ \).

Step3: Use angle bisector property

\( \overrightarrow{EB} \) bisects \( \angle AEC \), so \( m\angle AEB=\frac{1}{2}m\angle AEC \). Substituting \( m\angle AEC = 90^\circ \), we get \( m\angle AEB = \frac{90^\circ}{2}=45^\circ \).

For the first blank (measure of \( \angle CED \)): A right angle is \( 90^\circ \), so the answer is 90.
For the angle related to \( \angle CED \) in the straight angle: \( \angle AEC \) (since \( \angle AED=\angle AEC + \angle CED \) and \( \angle AED \) is straight, \( \angle AEC = 90^\circ \) as shown).

Answer:

First blank (measure of \( \angle CED \)): 90
Angle in the dropdown (related to \( \angle CED \) in \( \angle AED \)): \( \angle AEC \)