Question

In the adjoining figure, $ABCD$ is a parallelogram. If $P$ and $Q$ are points on $AD$ and $BC$ respectively such that $AP=\frac{1}{3}AD$ and $CQ=\frac{1}{3}BC$, prove that $AQCP$ is a parallelogram.

Answer 1

We have: $\angle B=\angle D$ [Opposite angles of parallelogram $ABCD$]

$AD=BC$ and $AB=DC$ [Opposite sides of parallelogram $ABCD$]
Also, $AD\parallel BC$ and $AB\parallel DC$
It is given that $AP=\frac{1}{3}AD$ and $CQ=\frac{1}{3}BC$
$\therefore AP=CQ$ [$\because AD=BC$ ]

$DP=\frac{2}{3}AD$ and $QB=\frac{2}{3}BC$

$\therefore DP=QB[\because AD=BC]$

In $△DPC$ and $△BQA$ we have:
$AB=CD$ [Opposite sides of parallelogram]

$\angle B=\angle D$ [Opposite angles of parallelogram]

$DP=QB$ [proved above]

$△DPC\cong △BQA$ [By SAS congruence Rule]

​ $\therefore$ $PC=QA$ [CPCT]
Thus, in quadrilateral $AQCP$, we have:
$AP=CQ$
$PC=QA$

Thus, in quadrilateral $AQCP$ opposite sides are equal.

$\therefore AQCP$ is a parallelogram.