Question

In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP$=$BQ. Show that APCQ is a parallelogram.

Answer 1

Given $ABCD$ is parallelogram and $DP=BQ$

In ${△}^{le}ABQ\&{△}^{le}CDP,$

$AB=CD$ ( opposite sides of $\parallel gm$)

$\Rightarrow \measuredangle ABQ=\measuredangle CDP$ ( alt. int. angles )

$DP=BQ$ ( given )

$\therefore {△}^{le}ABQ\cong △CDP$ By SAS congrency

$\therefore AQ=CP$ ( c.p.c.t ) $⟶\left(1\right)$

$\Rightarrow 180-\measuredangle AQB=180-\measuredangle CPD$

$\measuredangle AQP=\measuredangle CPQ$

$AQ\parallel CP⟶\left(2\right)$

Again, In ${△}^{le}APB\&{△}^{le}CQB,$

$AD=CB$ ( opposite sides of $\parallel gm$)

$\Rightarrow \measuredangle ADP=\measuredangle CBQ$ ( alt. int. angles )

$DP=BQ$ ( given )

$\therefore {△}^{le}APD\cong △CQB$ By SAS congrency

$\therefore AP=CQ$ ( c.p.c.t ) $⟶\left(3\right)$

$\Rightarrow 180-\measuredangle APD=180-\measuredangle CQB$

$\measuredangle APQ=\measuredangle CQP$

$AB\parallel CQ⟶\left(4\right)$

From $(1⟨\left(2\right),\left(3\right)\&\left(4\right)$

$APCQ$ is a parallelogram.