Question

A point P is taken on side CD of a parallelogram ABCD and CD is produced to Q making DQ = CP. The line through Q parallel to AD meets BP produced at S and AD is produced to meet BS at R. Prove that ARSQ is a parallelogram.

Answer 1

$QD=CP$ [ Given ] ---- ( 1 )

$AB=CD$ [ Opposite sides of parallelogram ]

$\Rightarrow$ $AB=CP+DP$

$\Rightarrow$ $AB=QD+DP$ [ From ( 1 ) ]

$\Rightarrow$ $AB=QP$

$\therefore$ $AB\parallel QP$ [ Since, $AB\parallel CD$ ]

$\therefore$ $ABPQ$ is a parallelogram.

$\therefore$ $AQ\parallel BP$

As $SR$ is extended part of $BP$

$\Rightarrow$ $AQ\parallel SR$

$\Rightarrow$ $AQ\parallel RP$

$\Rightarrow$ $QS\parallel AR$

$\therefore$ $ARSQ$ is a parallelogram.