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Given:  AD is the median of ΔABC. E is the mid point of AD. BE produced meets AD at FTo Prove : Construction: From Point D, draw DG BF.Proof:  In ΔADG, E is the mid-point of AD and EF||DG.∴F is the mid point of AG  [Converse of the mid point theorem]⇒ AF = FG  ... (i)In ΔBCF, D is the mid point of BC and DG||BF∴ G is the mid point of CF⇒ FG = GC  ... (ii)From (i) and (ii), we get,AF = FG = GC  ... (iii)Now, AF + FG + GC = AC⇒ AF + AF + AF = AC  [Using (iii)]⇒ 3AF = AC  
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