2009 AMC 10A Problem 17

The diagonal is $DB = \sqrt{4^2 + 3^2} = 5.$ Right triangles $EBA,$ $DBC,$ and $BFC$ are all similar to $\triangle DBA.$ From $\triangle EBA,$ $$\begin{aligned} \dfrac{EB}{AB} &= \dfrac{DB}{BC} \\ &\implies \dfrac{EB}{4} = \dfrac{5}{3} \\ &\implies EB = \dfrac{20}{3}. \end{aligned}$$ From $\triangle BFC,$ $$\begin{aligned} \dfrac{BF}{BC} &= \dfrac{DB}{AB} \\ &\implies \dfrac{BF}{3} = \dfrac{5}{4} \\ &\implies BF = \dfrac{15}{4}. \end{aligned}$$ Therefore $$\begin{aligned} EF &= EB + BF \\ &= \dfrac{20}{3} + \dfrac{15}{4} \\ &= \dfrac{125}{12}. \end{aligned}$$ Thus, the correct answer is C.