The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle if

1. PQRS is a parallelogram

2. PQRS is a rectangle

3. the diagonals of PQRS are perpendicular to each other

4. the diagonals of PQRS are equal.

Let ABCD be a quadrilateral with P, Q, R and S as mid-points of AB, BC, CD and DA respectively.

Let PR ⊥ QS.

In △QRP,

A and B are midpoints of PQ and QR respectively.

∴ AB || PR and AB = $\dfrac{1}{2}PR$ (By midpoint theorem) ……..(1)

Similarly in △PRS,

D and C are midpoints of PS and RS respectively.

∴ DC || PR and DC = $\dfrac{1}{2}PR$ (By midpoint theorem) ……..(2)

In △PQS,

D and A are midpoints of PS and PQ respectively.

∴ DA || QS and DA = $\dfrac{1}{2}QS$ (By midpoint theorem)………(3)

Similarly in △QRS,

B and C are midpoints of QR and SR respectively.

∴ BC || QS and BC = $\dfrac{1}{2}QS$ (By midpoint theorem)………(4)

From 1 and 2 we get,

AB = DC and AB || DC

From 3 and 4 we get,

DA = BC and DA || BC

Hence, proved that ABCD is a parallelogram.

Since, DA || QS and PR ⊥ QS

∴ DA ⊥ PR.

Since, DC || PR and PR ⊥ QS

∴ DC ⊥ QS.

In OMDN,

∠OMD = ∠SMC (Vertically opposite angle are equal)

In quadrilateral sum of angles = 360°

∠O + ∠M + ∠N + ∠D = 360°

90° + 90° + 90° + ∠D = 360°

∠D = 360° - 270° = 90°.

In parallelogram sum of alternate angles = 180°.

∠D + ∠B = 180°

90° + ∠B = 180°

∠B = 90°

Since, opposite sides are equal and adjacent sides are perpendicular to each other.

Hence, ABCD is a rectangle.

The quadrilateral formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a rectangle if the diagonals of PQRS are perpendicular to each other.

Hence, Option 3 is the correct option.