Question

6. in the figure ,ED AND EA are opposite rays ,and EB bisects

Answer 1

Since:

EB bisects ∠AEC

`m\angle AEB=m\angle CEB`

so:

`\begin{gathered} m\angle AEB+m\angle CEB+m\angle CED=180 \\ (7x-2)+(7x-2)+72=180 \\ 14x+68=180 \\ 14x=180-68 \\ 14x=112 \\ x=(112)/(14) \\ x=8 \end{gathered}`

Therefore:

`\begin{gathered} m\angle BEC=7(8)-2 \\ m\angle BEC=54 \end{gathered}`