Since D and E are the mid-points of sides BC and AC of ΔABC, therefore,
DE || BA and DE = \(\frac12\) BA (By mid-point theorem)
⇒ DE || FA and DE = FA (∵ FA = \(\frac12\) BA)
Also, DF || AC and DF = \(\frac12\) AC (By mid-point theorem)
⇒ DF || AE and DF = AE.
∴ DEAF is a parallelogram, whose diagonals AD and FE intersect at P.
Since the diagonals of a parallelogram bisect each other, therefore, AP = PD and FP = PE
⇒ P is the mid-point of FE
⇒ DP is the median of ΔDEF.
Similarly it can be shown that, FDEC is a parallelogram and hence R is the mid-point of ED and hence FR is the median of ΔDEF.
∴ Medians DP and FR intersect at point G, where G is the centroid of ΔDEF.
