Question

9. ABCD is a square. E, F, G and Hare the mid points of AB, BC, CD and DA respectively.
Such that AE=BF=CG=DH. Prove that EFGH is a square.
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Answer 1

Answer:

AE = BF = CG = DH = x (say)

BE = CF = DG = AH = y (say)

In ΔAEH and ΔBEF, we have

AE = BF

∠A = ∠B

And

AH = BE

So, by SAS congruency criterion,

we have

ΔAEH ≃ ΔBFE

∠1 = ∠2 and

∠3 = ∠4

But ∠1 + ∠3 = 90°

and ∠2 + ∠A = 90°

∠1 + ∠3 + ∠2 + ∠A = 90° + 90°

∠1 + ∠4 + ∠1 + ∠4 = 180°

2(∠1 + ∠4) = 180°

∠1 + ∠4 = 90°

HEF = 90°

Similarly we have ∠F = ∠G = ∠H = 90°

Hence, EFGH is a Square.

Step-by-step explanation:

Answer 2

Step-by-step explanation:

I hope this is right answer