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

1. ABCD is a parallelogram

2. ABCD is a rhombus

3. the diagonals of ABCD are equal

4. the diagonals of ABCD are 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

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.

∴ The quadrilateral formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order is a rhombus if the diagonals of ABCD are equal.

Hence, Option 3 is the correct option.