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Asked by ashwati 8th March 2009, 5:58 PM

Consider a quadrilateral ABCD with mid points of AB,BC,CD and DA to be P,Q,R  and S.

join the mid points to get quadrilateral PQRS.

Join diagonal AC of rectangle ABCD.

In triangle DAC,

SR parallel to AC and half of it( mid pt thm)

Simly,

PQ parallel to AC and half of it

So PQRS is a parallelogram( one pair of oposite sides is parallel and equal)

Now to show that it's a rhombus,

we can eitther show that adjacent sides are equal or that the diagonals of PQRS bisect each other at 90 degrees.

We'll go for the first option.

and

AB=DC=2y(say)

So,

AS=BQ=x

and

AP=PB=y

We know that the angles of a rectangle are 90 degrees each.

So, using Pythagoras' them,

we get,PS=

Simly PQ=

So we see that,

PQ=PS

So,

PQRS  is a parallelogram with adjacent sides equal.

So PQRS  must be a rhombus

Answered by Expert 8th March 2009, 10:19 PM
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