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

The correct option is D: diagonals of ABCD are equal.

Given: ABCD is a quadrilateral in which P, Q, R, and S are the midpoints of AB, BC, CD, and DA respectively. And PQRS is a rhombus.

In triangle ABD, we have

P and Q are the midpoints of AB and AD respectively.

$\therefore PS\parallel BD$ and $PS=\frac{1}{2}BD\dots \dots \left(i\right)$ [By midpoint theorem]

In triangle BCD, we have

R and Q are the midpoints of CD and BC respectively.

$\therefore RQ\parallel BD,RQ=\frac{1}{2}BD\dots \dots \left(ii\right)$ [By midpoint theorem]

In triangle ADC, we have

S and R are the midpoints of AD and CD respectively.

$\therefore SR\parallel AC$ and $SR=\frac{1}{2}AC\dots \dots \left(iii\right)$ [By midpoint theorem]

In triangle ABC, we have

P and Q are the midpoints of AB and BC respectively.

$\therefore PQ\parallel AC$ and $PQ=\frac{1}{2}AC\dots \dots \left(iv\right)$ [By midpoint theorem]

PQRS is a rhombus.

so, $PQ=QR=RS=PS\dots \dots \left(v\right)$

Therefore, $AC=BD$ [Using eq.(i), (ii), (iii), (iv), (v)]

Hence, the diagonal of ABCD are equal.