The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if

1. ABCD is a rhombus

2. diagonals of ABCD are equal

3. diagonals of ABCD are perpendicular to each other

4. diagonals of ABCD are equal and perpendicular to each other.

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

Let diagonals be of equal length i.e. AC = BD = x and AC ⊥ BD.

In △BCA,

P and Q are midpoints of AB and BC respectively.

∴ PQ || AC and PQ = $\dfrac{1}{2}AC = \dfrac{1}{2}x$ (By midpoint theorem) ……..(1)

Similarly in △ACD,

S and R are midpoints of AD and CD respectively.

∴ SR || AC and SR = $\dfrac{1}{2}AC = \dfrac{1}{2}x$ (By midpoint theorem) ……..(2)

In △ABD,

S and P are midpoints of AD and AB respectively.

∴ SP || BD and SP = $\dfrac{1}{2}BD = \dfrac{1}{2}x$ (By midpoint theorem)………(3)

Similarly in △BCD,

Q and R are midpoints of BC and CD respectively.

∴ QR || BD and QR = $\dfrac{1}{2}BD = \dfrac{1}{2}x$ (By midpoint theorem)………(4)

From 1, 2, 3 and 4 we get,

PQ = SR = SP = QR.

Hence, proved that PQRS is a rhombus.

Since, SP || BD and AC ⊥ BD

∴ SP ⊥ AC.

Since, SR || AC and AC ⊥ BD

∴ SR ⊥ BD.

In OMSN,

∠OMS = ∠DMR (Vertically opposite angle are equal)

In quadrilateral sum of angles = 360°

∠O + ∠M + ∠N + ∠S = 360°

90° + 90° + 90° + ∠S = 360°

∠S = 360° - 270° = 90°.

Since, in rhombus adjacent angles sum = 180°

∠S + ∠R = 180°

90° + ∠R = 180°

∠R = 90°.

∠Q + ∠R = 180°

90° + ∠Q = 180°

∠Q = 90°.

∠S + ∠P = 180°

90° + ∠P = 180°

∠P = 90°.

Since, PQ = QR = RS = SP and ∠P = ∠Q = ∠R = ∠S = 90°

Hence, proved that PQRS is a square.

The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if diagonals of ABCD are equal and perpendicular to each other.

Hence, Option 4 is the correct option.