It is given that AK = BL = CM = DN
ABCD is a square
So we get
BK = CL = DM = AN …… (1)
Consider △ AKN and △ BLK
It is given AK = BL
From the figure we know that ∠ A = ∠ B = 90o
Using equation (1)
AN = BK
By SAS congruence criterion
△ AKN ≅ △ BLK
We get
∠ AKN = ∠ BLK and ∠ ANK = ∠ BKL (c. p. c. t)
We know that
∠ AKN + ∠ ANK = 90o
∠ BLK + ∠ BKL = 90o
By adding both the equations
∠ AKN + ∠ ANK + ∠ BLK + ∠ BKL = 90o + 90o
On further calculation
2 ∠ ANK + 2 ∠ BLK = 180o
Dividing the equation by 2
∠ ANK + ∠ BLK = 90o
So we get
∠ NKL = 90o
In the same way
∠ KLM = ∠ LMN = ∠ MNK = 90o
Therefore, it is proved that KLMN is a square.