Question

Show that the line joining the mid-points of the consecutive sides of a quadrilateral from a parallelogram.

Answer 1

In $△ADC,$

$S$ is the mid-point of $AD$ and $R$ is the mid-point of $CD$

$\therefore SR\parallel AC$ and $SR=\frac{1}{2}AC$ ....$\left(1\right)$ since line segments joining the mid-points of two sides of a triangle is parallel to the third side and is half of it.

In $△ABC,$

$P$ is the mid-point of $AB$ and $Q$ is the mid-point of $BC$

$\therefore PQ\parallel AC$ and $PQ=\frac{1}{2}AC$ ....$\left(2\right)$ since line segments joining the mid-points of two sides of a triangle is parallel to the third side and is half of it.

From $\left(1\right)$ and $\left(2\right)$

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

So, in $PQRS$, one pair of opposite sides are parallel and equal.

Hence $PQRS$ is a parallelogram.

$PR$ and $SR$ are the diagonals of parallelogram $PQRS$

So, $OP=OR$ and $OQ=OS$ (Diagonals of a parallelogram bisect each other)

Hence proved.