jmc

First of all, let us label the tip of the triangle. Since $MN\parallel AB,$ we know that $\angle CMN=\angle CAB$ and $\angle CNM=\angle CBA.$ Therefore, by AA similarity, we have $△ABC\sim MNC.$ Then, we find: $$\begin{array}{rl}\frac{AC}{MC}& =\frac{BC}{NC}\\ \frac{AM+MC}{MC}& =\frac{BN+NC}{NC}\\ 1+\frac{AM}{MC}& =1+\frac{BN}{NC}\\ \frac{5}{3}& =\frac{BN}{2.4}.\end{array}$$ Therefore, $BN=\frac{5\cdot 2.4}{3}=\boxed{4}.$