Question

in the figure, $overrightarrow{ed}$ and $overrightarrow{ea}$ are opposite rays, and $overrightarrow{eb}$ bisects $angle aec$. if $mangle ced = 56^{circ}$ and $mangle aeb=(7x - 8)^{circ}$, then what is $mangle bec? mangle bec=square^{circ}$

Explanation:

Step1: Find m∠AEC

Since $\overrightarrow{ED}$ and $\overrightarrow{EA}$ are opposite rays, $\angle AED = 180^{\circ}$. Given $m\angle CED=56^{\circ}$, then $m\angle AEC = 180^{\circ}-m\angle CED$. So $m\angle AEC=180 - 56=124^{\circ}$.

Step2: Use the angle - bisector property

Since $\overrightarrow{EB}$ bisects $\angle AEC$, then $m\angle AEB=m\angle BEC$. And we know that $m\angle AEB=(7x - 8)^{\circ}$ and $m\angle AEC = 124^{\circ}$, also $m\angle AEC=2m\angle AEB$. So $2(7x - 8)=124$.
First, expand the left - hand side: $14x-16 = 124$.
Add 16 to both sides: $14x=124 + 16=140$.
Divide both sides by 14: $x = 10$.
Then $m\angle AEB=7x-8=7\times10 - 8=62^{\circ}$.
Since $m\angle BEC=m\angle AEB$, $m\angle BEC = 62^{\circ}$.

Answer:

$62$