Question

in rectangle ABCD, E ; F ; G ; H are points on sides AB ; BC ; CD ; DA respectively. AE=CG AND BF=DH. prove that EPGH is a parallelogram.

Answer 1

In triangle AHE AND FGC ...1.AE = GC(GIVEN)
2.ANGLE HAE= ANGLE FCG = 90 DEGREE( ALL ANGLES OF RECTANGLE R 90 DEGRESS)
3.ANGLE EHA = ANGLE CGF = 45 DEGREE( ANGLE  SUM PROPERTY OF TRIANGLE)
So by AAS axion of congruency AHE is congruent to  FGC
so by cpct EH = GF 
therefore EH is parallel to GF
Similarly in Triangle BEF AND TRIANGLE HDG
1. HD = BF(GIVEN)
2. ANGLE HDG = ANGLE EBF = 90 DEGREES
3.ANGLE DGH = ANGLE BFE = 45 DEGREES
So by AAS axiom of congruency 
 Triangle BEF is congruent to TRIANGLE HDG
so by cpct HG = EF
therefore HG is parallel toEF
And hence 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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