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given: m∠aeb = 45° ∠aec is a right angle. prove: (overrightarrow{eb}) b…

Question

given: m∠aeb = 45°
∠aec is a right angle.
prove: (overrightarrow{eb}) bisects ∠aec.
proof: we are given that m∠aeb = 45° and ∠aec is a right angle. the measure of ∠aec is 90° by the definition of a right angle. applying the
blank gives m∠aeb + m∠bec = m∠aec. applying the substitution property gives 45°+ m∠bec = 90°. the subtraction property can be used to find m∠bec = 45°, so ∠bec ≅ ∠aeb because they have the same measure. since (overrightarrow{eb}) divides ∠aec into two congruent angles, it is the angle bisector.

Explanation:

Step1: Recall right - angle definition

Since $\angle AEC$ is a right angle, $m\angle AEC = 90^{\circ}$.

Step2: Use angle - addition postulate

By the angle - addition postulate, $m\angle AEB+m\angle BEC=m\angle AEC$.

Step3: Substitute known values

We know $m\angle AEB = 45^{\circ}$ and $m\angle AEC = 90^{\circ}$, so $45^{\circ}+m\angle BEC = 90^{\circ}$.

Step4: Solve for $m\angle BEC$

Using the subtraction property of equality, $m\angle BEC=90^{\circ}- 45^{\circ}=45^{\circ}$.

Step5: Check for congruent angles

Since $m\angle AEB = 45^{\circ}$ and $m\angle BEC = 45^{\circ}$, $\angle AEB\cong\angle BEC$.

Step6: Apply angle - bisector definition

Since $\overrightarrow{EB}$ divides $\angle AEC$ into two congruent angles, $\overrightarrow{EB}$ bisects $\angle AEC$.

Answer:

$\overrightarrow{EB}$ bisects $\angle AEC$ is proven.