Question

In given figure $ABCD$ 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:
$SR\parallel AC$ and $SR=\frac{1}{2}AC$

Answer 1

(i) In $△ACD$, we have $S$ is the mid-point of $AD$ and $R$ is the mid- point of $CD$.

Then $SR\parallel AC$

Using Mid point theorem $SR=\frac{1}{2}AC$

(ii) In $△ABC$,

$P$ is the mid-point of the side $AB$ and $Q$ is the mid-point of the side $BC$.
Then, $PQ\parallel AC$
and using Mid point Theorem

$PQ=\frac{1}{2}AC$
Thus, we have proved that :

$PQ\parallel AC$ and $SR\parallel AC$

$\Rightarrow PQ\parallel SR$
Also $PQ=SR=\frac{1}{2}AC$

(iii) Since $PQ=SR$ and $PQ\parallel SR$

One pair of opposite sides are equal and parallel.

$\Rightarrow PQRS$ is a parallelogram.