Given,

ABCD is a square

E,F,G,H are the points of AB, BC, CD, DA

Such that AE = BF = CG = DH

In ΔAEH & ΔBFE

Let,

AE = BF = CG = DH = x

BE = CF = DG = AF = y

In ΔAEH & ΔBFE

AE = BF (given)

∠A = ∠B (each equal)

AH = BE

So, by SAS congruency

ΔAEH ≅ΔBFE

∠1 = ∠2 & ∠3 = ∠4

∠1 + ∠3 = 90^∘

∠2 + ∠4 = 90^∘

∠1+∠2+∠3+∠4 = 180^∘

∠1+∠4+∠1+∠4 = 180^∘

2(∠1+∠4) = 180^∘

∠1+∠4 = = 90^∘

So, ∠HEF = 90^∘

Similarly we have,

∠F =∠G = ∠H = 90^∘

Hence, EFGH is a square.