Question

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.

Answer 1

Here, $ABCD$ is a quadrilateral. $AC$ and $BD$ are diagonals, which are perpendicular to each other.

$P,Q,R$ and $S$ are the mid-point of $AB,BC,CD$ and $AD$ respectively.

In $△ABC,$

$P$ and $Q$ are mid points of $AB$ and $BC$ respectively.

$\therefore$ $PQ\parallel AC$ and $PQ=\frac{1}{2}AC$ ----- ( 1 ) [ By mid-point theorem ]

Similarly, in $△ACD,$

$R$ and $S$ are mid-points of sides $CD$ and $AD$ respectively.

$\therefore$ $SR\parallel AC$ and $SR=\frac{1}{2}AC$ ----- ( 2 ) [ By mid-point theorem ]

From ( 1 ) and ( 2 ), we get

$PQ\parallel SR$ and $PQ=SR$

$\therefore$ $PQRS$ is a parallelogram.

Now, $RS\parallel AC$ and $QR\parallel BD.$

$\Rightarrow$ Also, $AC\perp BD$ [ Given ]

$\therefore$ $RS\perp QR$

$\therefore$ $PQRS$ is a rectangle.