Question

$ ABCD$ is a quadrilateral in which $ P, Q, R$ and $ S$ are mid-points of the sides $ AB, BC, CD$ and $ DA$. $ AC$ is a diagonal. Show that

Answer 1

Step $1:$ Proving $SR\parallel AC$ and $SR=\frac{1}{2}AC$ :

In, $∆ADC$,

$S$ is the midpoint of $AD$ and $R$ is the midpoint of $CD$.

Therefore, by mid point theorem, the line segment joining the mid-points of two sides of a triangle is parallel to the third side and half ot it.

Therefore, $SR\parallel AC$ and

$SR=\frac{1}{2}AC$…………………..(i)

Step $2:$ Proving $PQ=SR$:

In, $∆ABC$,

$P$ is the midpoint of $AB$ and $Q$ is the midpoint of $BC$.

Therefore, by mid point theorem, the line segment joining the mid-points of two sides of a triangle is parallel to the third side and half ot it.

Therefore, $PQ\parallel AC$ and

$PQ=\frac{1}{2}AC$……………..(ii)

From (i) and (ii), we get

Therefore, $PQ=SR$

Step $3:$Proving $PQRS$ is a parallelogram:

We have, $SR\parallel AC$ and

$PQ\parallel AC$

$\therefore SR\parallel PQ$ (Lines parallel to same line are parallel to each other)

And, also $PQ=SR$.

$\therefore PQRS$ is a parallelogram because a pair of opposite side of quadrilateral $PQRS$ is equal and parallel.

Hence proved.