Draw DG || BF In ΔADG,E is the midpoint of AD and EF || DG ∴ F is the mid-point of AG. (Converse of mid-point theorem) ⇒ AF = FG …..(1) In Δ BFC, D is the mid-point of BC and DG||BF.∴ G is the mid-point of FC. (Converse of mid-point theorem) ⇒ FG = GC......(2) From (a) and (b), we have AF = FG = GC ∴AF=