Given: FB = 3 cm, AF = 7 cm and EF = 4 cm.From the diagram, E is a point on BC and line segment DF is drawn where F is on extended side AB.In ∆ FBE and ∆ FADSince AD || BC, ∠FBE = ∠FAD (corresponding angles) and∠FEB = ∠FDA (corresponding angles).Thus, ∆ FBE ∼ ∆ FADBy similarity of ∆ FBE ∼ ∆ FAD, the ratios of corresponding sides are equal$\;\frac{FB}{FA} = \;\frac{FE}{FD}$$\;\frac{3}{7} = \;\frac{4}{FD}$$3 \times FD = 7 \times 4$$3 \times FD = 28$$FD = \;\frac{28}{3} \text{ cm}$The length FD is $\;\frac{28}{3}\mathrm{cm}$.