# 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 ΔABC

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=12AC

In ΔADC

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=12AC

PQ||RS and PQ=RS

So, PQRS is a parallelogram. [one pair of opposite side is parallel and equal]

In ΔAPS & ΔBPQ

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

∠PAS=∠PBQ(All the angles of rectangle are 90o)

AS=BQ

∴ΔAPS≅ΔBPQ(SAS congruency)

∴PS=PQ

PS=RQ & PQ=RS (opposite sides of parallelogram is equal)

PQ=RS=PS=RQ[All sides are equal]

PQRS is a parallelogram with all sides equal

∴ So PQRS is a rhombus.