ABCD is a square. E, F, G and H are points on AB, BC, CD and DA respectively,

Solution:

We have,

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.