We know that if [tex]m\angle ABD=2(m\angle DBC)[/tex], then [tex]m\angle ABD=\frac{2n}{3}[/tex] and [tex]m\angle DBC=\frac{n}{3}[/tex].
Since BE bisects [tex]\angle DBC[/tex], we know that [tex]\angle DBE \cong \angle EBC[/tex], and thus [tex]\angle EBC=\boxed{\frac{n}{6}}[/tex]
