Question

In the given figure, ABCD is a quadrilateral whose diagonals intersect at right angles. Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides is a rectangle.

Answer 1

Given ABCD is a quadrilateral and P,Q,R and S be the midpoints of AB, BC, CD and DA respectively.
AC and BD are the diagonals which intersect each other at O.
RQ intersects AC at E and SR intersects BD at F.

In $\Delta ABC$, we have:
$\therefore PQ\mid \mid AC$ and $PQ=\frac{1}{2}AC$ [By midpoint theorem]
Again, in $\Delta DAC$, the points S and R are the midpoints of AD and DC, respectively.
$\therefore SR\mid \mid AC$ and $SR=\frac{1}{2}AC$ [By midpoint theorem]

Now, $PQ\mid \mid AC$ and $\Rightarrow PQ\mid \mid SR$
Also, $PQ=SR$, [Each equal to $\frac{1}{2}AC$ ] ...(i)
So, PQRS is a parallelogram.

We know that the diagonals of the given quadrilateral bisect each other at right angles.
$\therefore \angle EOF={90}^{\circ }$
​Now $RQ\mid \mid DB$,
$\Rightarrow RE||FO$
Also,$SR||AC$
$\Rightarrow FR||OE$
$\therefore OERF$ is a parallelogram.
So,$\angle FRE=\angle EOF={90}^{\circ }$ (Opposite angles are equal)
Thus, PQRS is a parallelogram with $angle\angle R={90}^{\circ }$.
Therefore, ​ PQRS is a rectangle.