Question

ABCD is a square E, F, G and H are points on AB, BC, CD and DA respectively, such that AE=BF=CG=DH. Prove that EFGH is a square.

Answer 1

EFGH is square

Step-by-step explanation:

Let say side of Square = a

AB = BC = CD = DA = a

Let say AE=BF=CG=DH = b

=> BE = CF = DG = AH = (a - b)

=> EF² = FG² = GH² = EH² = b² + (a - b)²

all sides are equal

Tan∠AEH =  (a - b)/b

Tan ∠BEF = b/(a - b)

Tan (∠AEH + ∠BEF )  =   (Tan∠AEH  + Tan ∠BEF )/(1 - (Tan∠AEH *Tan∠BEF))

=> Tan (∠AEH + ∠BEF )  =  ((a - b)/b + b/(a - b))/(1 - (((a - b)/b) * )b/(a - b)))

=>  Tan (∠AEH + ∠BEF )  =  ((a - b)/b + b/(a - b))/(1 - 1 )

=>  Tan (∠AEH + ∠BEF )  =  ((a - b)/b + b/(a - b))/0

=> Tan (∠AEH + ∠BEF )  =  ∞

=> ∠AEH + ∠BEF = 90°

∠AEH + ∠HEF + ∠BEF = 180°   ( Straight line)

=> ∠HEF  + ∠AEH + ∠BEF = 180°

=> ∠HEF  + 90° = 180°

=> ∠HEF = 90°

Similarly we can show all other angles are 90°

All sides of EFGH are equal

& all angles = 90°

Hence EFGH is square

QED

Proved

Learn more:

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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