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

Correct option is C

Given:

Quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS

Solution:

Let us join AC and BD

In triangle ABC

P and Q are the mid points of AB, BC respectively

So, $$PQ \parallel AC$$​ and PQ =$$\frac{1}{2}$$​AC   ….(1)

Similarly in triangle ADC

​$$SR \parallel AC$$ and SR = $$\frac{1}{2}$$​AC    …(2)

Clearly $$PQ \parallel SR$$​ and PQ=SR

Since in quadrilateral PQRS one pair of side is equal and parallel to each parallelogram

So $$PS \parallel QR$$​ and PS = QR     ….(3)

In triangle BCD Q and R are mid point of side BC and CD respectively

So $$QR \parallel BD$$​ and QR = $$\frac{1}{2}$$​ BD   ….(4)

However the diagonal of rectangle is equal

By using eq(1),(2),(3) and (4)

PQRS is a rhombus if AC and BD are equal