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Let G be the centroid of triangle ABC. Given E and F are the mid points of BC and AC respectively. Thus, by mid point theorem, AD∥EF
AB=2EF
AD=EF (I) (D is mid point of AB)
Now, In △ADG and △GEF,
∠AGD=∠EGF (Vertically opposite angles)
AD=EF (From I)
∠ADG=∠GFE (Alternate angles for parallle lines EF and AD)
△ADG≅△EGF (ASA rule)
Thus, AG=GE (Corresponding sides)
Also, DG=GF (Corresponding sides)
Thus, AE and DF bisect each other at G.
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