Let ABCD is a square such that AB = BC = CD = DA, AC = BD and P, Q, R and S are the mid points of the sides AB, BC, CD and DA respectively.

In

P and Q are the mid-points of AB and BC respectively.

Therefore,

PQ ‖ AC and PQ = AC (Mid-point theorem) (i)

Similarly,

In

SR ‖ AC and SR = AC (Mid-point theorem) (ii)

Clearly,

PQ ‖ SR and PQ = SR

Since, in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other. Hence, it is a parallelogram.

Therefore,

PS ‖ QR and PS = QR (Opposite sides of a parallelogram) (iii)

In

Q and R are the mid-points of sides BC and CD respectively

Therefore,

QR ‖ BD and QR = BD (Mid-point theorem) (iv)

However, the diagonals of a square are equal

Therefore,

AC = BD (v)

By using equation (i), (ii), (iii), (iv) and (v), we obtain

PQ = QR = SR = PS

We know that, diagonals of a square are perpendicular bisector of each other

Therefore,

∠AOD = ∠AOB = ∠COD = ∠BOC = 90^o

Now, in quadrilateral EHOS, we have

SE || OH

Therefore,

∠AOD + ∠AES = 180^o (Corresponding angle)

∠AES = 180° - 90°

= 90°

Again,

∠AES + ∠SEO = 180^o (Linear pair)

∠SEO = 180° - 90°

= 90°

Similarly,

SH || EO

Therefore,

∠AOD + ∠DHS = 180^o (Corresponding angle)

∠DHS = 180° - 90° = 90°

Again,

∠DHS + ∠SHO = 180° (Linear pair)

∠SHO = 180° - 90°

= 90°

Again,

In quadrilateral EHOS, we have

∠SEO = ∠SHO = ∠EOH = 90°

Therefore, by angle sum property of quadrilateral in EHOS, we get

∠SEO + ∠SHO + ∠EOH + ∠ESH = 360°

90^o + 90^o + 90^o + ∠ESH = 360°

∠ESH = 90°

In the same manner, in quadrilateral EFOP, FGOQ, GHOR, we get

∠HRG = ∠FQG = ∠EPF = 90°

Therefore, in quadrilateral PQRS, we have

PQ = QR = SR = PS and ∠ESH = ∠HRG = ∠FQG = ∠EPF = 90°

Hence, PQRS is a square.