please answer this question

Asked by  | 3rd Jan, 2009, 01:25: PM

Expert Answer:

Draw a quadrilateral ABCD.

Mark the mid points P,Q,R  and S as given in the question.

Now,

 PS is parallel to and half of BD.

Simly, QR is parallel to and half of  BD.

So, PS=QR

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

Remember that diagonals of a rhombus bisect each other at right angles?

So,angle AOB  is 90 degrees.

Mark the intersection of PS  with AC  as H  and PQ with BD  as G.

PS is parallel to BD , so PH  is parallel to GO.

Also,

PQ is parallel to AC, so PG is parallel to HO.

So, PGOH is a parallelogram (both pairs of opposite sides are parallel)

angle O is a right angle(proved before)

So , angle HPG is also a right angle(opposite angles of a parallelogram are equal)

Thus, PQRS is a parallelogram with one angle as a right angle, so by definition of a rectangle, PQRS becomes a rectangle.

 

Answered by  | 13th Jan, 2009, 03:54: PM

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