ABCD is a quadrilaterals E,F,G,H are mid points of AB,BC,CD and DA respectively prove that EFGH is a parallelogram
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Join the diagonal AD and BC
Now from mid point theorem, the line joining the mid points of two sides of a triangle is parallel to the third side
Therefore in triangle ABD, EH parallel to BD, and also in triangle BCD, FG parallel to BD
Therefore EH parallel to FG
Similarly in triangle ABC and ABC,EF parallel to GH
Therefore two pairs of opposite side are parallel
Therefore EFGH is a parallelogram
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