The diagonals of a quadrilateral intersect at right angles. Prove that the figure obtained by joining the mid-points of the adjacent sides of the quadrilateral is a rectangle.

Hint: In this question try to use the midpoint theorem. As per this statement, The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side.

Complete step-by-step answer:

Here, $ ABCD $ is a quadrilateral, $ AC $ and $ BD $ are diagonals which are perpendicular to each other.
 $ P,Q,R $ and $ S $ are the mid points of $ AB,BC,CD $ and $ DA $ respectively.
In triangle $ ABC $
 $ P $ and $ Q $ are the mid points of $ AB $ and $ BC $ respectively.
 $ \therefore PQ\parallel AC $ and $ PQ = \dfrac{1}{2}AC $ ....equation first (By mid point theorem)
Similarly in triangle $ ACD $
 $ R $ and $ S $ are the mid points of sides $ CD $ and $ AD $ respectively.
 $ \therefore SR\parallel AC $ and $ SR = \dfrac{1}{2}AC $ ....equation second (By mid point theorem)
From equation first and second we get
 $ PQ\parallel SR $ and $ PQ = SR $
 $ \therefore PQRS $ is a parallelogram
Now, $ RS\parallel AC $ and $ QR\parallel BD $
Also its given that $ AC \bot BD $
 $ \therefore RS \bot QR $
 $ \therefore PQRS $ is a rectangle.

Note: A rectangle is a quadrilateral in which opposites sides are parallel and equal, so above we proved that $ PQ\parallel SR $ and $ PQ = SR $ also in a rectangle interior angle is 90degree and above we proved that side $ RS \bot QR $ , hence $ PQRS $ is a rectangle.