Question

abcd is a quarilaterals E,F,G and H are mid points of AB. BC,CD and DA respectively. prove that EFGH is a parallelogram

Answer 1

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.

Answer 2

Given AE=BF=CG=DH

⟹ So, EB=FC=GD=HA

In △s AEH and BFE,

AE=BF, AH=EB,

∠A=∠B (each ∠ = 90⁰)

∴ △AEH ≅ △BFE

⟹ EH=EF and ∠4= ∠2.

But ∠1 + ∠4 = 90⁰ ⟹ ∠1 + ∠2 = 90⁰

⟹ ∠HEF = 90⁰

And if ∠HEF = 90⁰ so, ∠EFG = 90⁰, ∠FGH = 90⁰ and ∠GHE = 90⁰.

Hence Proved.

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