Since:
EB bisects ∠AEC
[tex]m\angle AEB=m\angle CEB[/tex]
so:
[tex]\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=\frac{112}{14} \\ x=8 \end{gathered}[/tex]
Therefore:
[tex]\begin{gathered} m\angle BEC=7(8)-2 \\ m\angle BEC=54 \end{gathered}[/tex]
