In the given figure, G is the point of concurrence of medians of Δ DEF. Take point H on ray DG such that D-G-H and DG = GH, then prove that `square` GEHF is a parallelogram.
G is the point of concurrence of the medians of Δ DEF.
Let the point where the median divides EF into two equal parts be A.
Thus, EA = AF. .....(1)
we know that the point of concurrence of the medians, divides each median in the ratio 2 : 1.
So, let DG = 2x and GA = x
Given that DG = GH
So, GA = AH = x
Thus, point A dividess EF and GH into two equal parts.
Hence , `square`GEHF is a parallelogram as the diagonals EF and GH bisect each other.