Concept: According to the midpoint theorem, "the line segment in a triangle joining the midpoints of two sides of the triangle is said to be parallel to its third side and also half the length of the third side."

Given: ABCDis a quadrilateral AC and  BD intersect at right angles.

P,Q,R,S are mid points of respective sides.

In $∆ABC$, $PQ\parallel AC$ and  $PQ=\frac{1}{2}AC$ (By mid point theorem)

In  $∆ADC$ , $SR\parallel AC$ , and $SR=\frac{1}{2}AC$ (By mid point theorem)

Therefore,  $PQ=SR$ and  $PQ\parallel SR$ similarly, using diagonal $DC,PS=RQ$ and 

hence,  PQRSis a parallelogram.  since,  $PQ\parallel AC$  and  $QR\parallel BD$

But,  $AC\perp BD$  Hence,  $PQ\perp QR$ (Angles between two lines = Angles between their parallels)

Thus,  PQRS  is a rectangle.