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In the adjoining figure, G is the point of concurrence of medians of ADEF. Take point H on ray DG such that D-G-H and DG = GH, then prove that GEHF is a parallelogram.

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Given: Point G (centroid) is the point of concurrence of the medians of ADEF. 

DG = GH 

To prove: □GEHF is a parallelogram. 

Proof:

Let ray DH intersect seg EF at point I such that E-I-F. 

∴ seg DI is the median of ∆DEF. 

∴ El = FI ……(i) 

Point G is the centroid of ∆DEF. 

∴ DG/GI = 2/1 [Centroid divides each median in the ratio 2:1]

∴ DG = 2(GI) 

∴ GH = 2(GI) [DG = GH]

∴ GI + HI = 2(GI) [G-I-H] 

∴ HI = 2(GI) – GI 

∴ HI = GI ….(ii) 

From (i) and (ii), 

GEHF is a parallelogram [A quadrilateral is a parallelogram, if its diagonals bisect each other]

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