Question

Abcd is a kite having ab equal ad and bc equal cd.prove that the figer formed by joining the mid point of the side in order is a rectangle

Answer 1

Given:A kite ABCD having AB=AD and BC=CD

P,Q,R,S are the midpoint of sides AB,BC,CD, and DA respectively.

PQ,QR,RS and spare joined

To prove: PQRS is a rectangle

Proof:In △ABC,P and Q are the midpoints of AB and BC respectively.

∴PQ∥AC and PQ=

2

1

AC ....(1)

In △ADC,R and S are the midpoint of CD and AD respectively.

∴RS∥AC and RS=

2

1

AC ....(2)

From (1) and (2) we have

PQ∥RS and PQ=RS

Thus, in quadrilateral PQRS, a pair of opposite sides are equal and parallel. So PQRS is a parallelogram.

Now, we shall prove that one angle of parallelogram PQRS it is a right angle.

Since AB=AD

⇒

2

1

AB=

2

1

AD

⇒AP=AS ........(3)

⇒∠1=∠2 ......(4)

Now,in △PBO and △SDR we have

PB=SD since AD=AB⇒

2

1

AD=

2

1

AAB

BQ=DR

∴PB=SD

And PQ=SR since PQRS is a parallelogram.

So by SSS criterion of congruence, we have

△PBQ≅△SOR

⇒∠3=∠4 by CPCT

Now, ∠3+∠SPQ+∠2=180

∘

And ∠1+∠PSR+∠4=180

∘

∴∠3+∠SPQ+∠2=∠1+∠PSR+∠4

⇒∠SPQ=∠PSR since ∠1=∠2 and ∠3=∠4

Now, transversal PS cuts parallel lines SR and PQ at S and P respectively.

∴∠SPQ+∠PSR=180

∘

⇒2∠SPQ=180

∘

⇒∠SPQ=

2

180

∘

=90

∘

Thus, PQRS is a parallelogram such that ∠SPQ=90

∘

Hence, PQRS is a parallelogram.

Hope this helps