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Question
In the given figure, G is the point of concurrence of medians of ΔDEF. 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
Let, the median side drawn from point D intersects EF at point M.
Point G is the concurrence point.
The concurrence point divides each median in the ratio 2:1.
∴ DG : GM = 2 : 1
∴ `("DG")/("GM") = 2/1`
∴ DG = 2GM ...(i)
∴ DG = GM + MH ...(G-M-H)
∴ 2GM = GM + MH ...[from (i)]
∴ 2GM – GM = MH
∴ GM = MH ...(ii)
In `square`GEHF,
Line GM ≅ Line MH ...[From (ii)]
Line EM ≅ Line MF ...(Point M is the midpoint of line EF)
A quadrilateral is a parallelogram if its diagonals bisect each other.
`square`GEHF is a parallelogram.
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