Answer
m$\angle$ABC = 108$^{\circ}$
Work Step by Step
Ray BD bisects $\angle$ABC, and rays BE and BF trisect $\angle$ABC
Given m$\angle$EBD = 18$^{\circ}$, we can conclude m$\angle$FBD is also equal to 18$^{\circ}$
m$\angle$EBF = m$\angle$FBD + m$\angle$EBD. Using substitution, m$\angle$EBF = 36$^{\circ}$
Since $\angle$EBF trisects $\angle$ABC, m$\angle$ABC = 3 * m$\angle$EBF
Using substitution, m$\angle$ABC = 3 * (36)
Finally, after simplifying the above statement, m$\angle$ABC = 108$^{\circ}$
