ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Here, we are joining A and C.

In $\Delta$ABC

P is the mid point of AB

Q is the mid point of BC

PQ$||$AC [Line segments joining the mid points of two sides of a triangle is parallel to AC(third side) and also is half of it]

PQ$=\frac{1}{2}$AC

In $\Delta$ADC

R is mid point of CD

S is mid point of AD

RS$||$AC [Line segments joining the mid points of two sides of a triangle is parallel to third side and also is half of it]

RS$=\frac{1}{2}$AC

So, PQ$||$RS and PQ$=$RS [one pair of opposite side is parallel and equal]

In $\Delta$APS & $\Delta$BPQ

AP$=$BP [P is the mid point of AB)

$\angle$PAS$=\angle$PBQ(All the angles of rectangle are $90^o$)

AS$=$BQ

$\therefore \Delta$APS$\cong \Delta$BPQ(SAS congruency)

$\therefore$PS$=$PQ

BS$=$PQ & PQ$=$RS (opposite sides of parallelogram is equal)

$\therefore$ PQ$=$RS$=$PS$=$RQ[All sides are equal]

$\therefore$ PQRS is a parallelogram with all sides equal

$\therefore$ So PQRS is a rhombus.