Question

ABCD is a quadrilateral; P,Q,R and S are the points of trisection of sides AB,BC,CD and DA respectively and are adjacent to A and C; prove thar PQRS is a parallelogram.

Answer 1

Given A quadrilateral ABCD in which P,Q,R and S are the points of trisection of sides AB,BC,CD and DA respectively and are adjacent to A and C.
To prove PQRS is a parallelogram i.e., PQ||SR and QR||PS.
Construction Join AC.
Proof Since P,Q,R and S are the points of trisection of AB,BC,CD and DA respectively.

$\therefore$ $BP=2PA,BQ=2QC,DR=2RC$ and, $DS=2SA$

In $△ADC$, we have

$\frac{DS}{SA}=\frac{2SA}{SA}=2$ and, $\frac{DR}{RC}=\frac{2RC}{RC}=2$

$\Rightarrow$ $\frac{DS}{SA}=\frac{DR}{RC}$

$\Rightarrow$ S and R divide the sides DA and DC respectively in the same ratio.

$\Rightarrow$ $SR||AC$ {By the converse of Thale's Theorem]

In $△ABC$, we have

$\frac{BP}{PA}=\frac{2PA}{PA}=2$ and $\frac{BQ}{QC}=\frac{2QC}{QC}=2$

$\Rightarrow$ $\frac{BP}{PA}=\frac{BQ}{QC}$

$\Rightarrow$ P and Q divide the sides BA and BC respectively in the same ratio.

$\Rightarrow$ $PQ||AC$ ........(ii)

From equations (i) and (ii), we have

$SR||AC$ amd $PQ||AC$

$\Rightarrow$ $SR||PQ$

Similarly, by joining BD, we can prove that

$\Rightarrow$ $QR||PS$

Hence, PQRS is a parallelogram.