Given : In ∆ABC, AD is the median and E is the mid point of AD. Construction : Draw DP parallel to EF To find = AF : FC Solution : in ∆ADP, E is the mid point of AD and EF ∥ DP. ⇒ F is mid point of AP. [By converse of mid point theorem] Similarly, in FBC, D is the mid point of BC and EF DP. ⇒ P is mid point of FC. Thus, AF = FP = PC ∴ AF=