Sum

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.

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#### Solution

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.

Concept: Tests for parallelogram

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