Let ABCD be the square and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively.

Join diagonals of the square.

In ∆ ABC, we have, by midpoint theorem,
∴ PQ ∣∣ AC and PQ = AC

Similarly, SR ∣∣ AC and SR = AC.

As, PQ ∣∣ AC and SR ∣∣ AC, then also PQ ∣∣ SR

Also, PQ = SR, each equal to AC …(1)

So, PQRS is a parallelogram

Now, in ∆SAP and ∆QBP, we have,

AS = BQ
∠A = ∠B = 90°
AP = BP

∴ By SAS test of congruency,

∆SAP ≅ ∆QBP​

Hence, PS = PQ …by cpct …(2)

Similarly, ∆SDR ≅ ∆QCR

∴ SR = RQ … by cpct …(3)

Hence, from 1, 2 and 3 we have,

PQ = PQ = SR = RQ

We know that the diagonals of a square bisect each other at right angles.
∴ ∠​EOF = 90^o
​Now, RQ ∣∣ DB
⇒RE ∣∣ FO
Also, SR ∣∣ AC
⇒FR ∣∣ OE
∴ OERF is a parallelogram.
So, ∠FRE = ∠EOF = 90^o​ (Opposite angles are equal)
Thus, PQRS is a parallelogram with ∠R = 90^o and PQ = PS = SR = RQ.

This means that PQRS is square.

Hence, the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a square is a square.