Question

In parallelogram $ABCD$, two point $P$ and $Q$ are taken on diagonal $BD$ such that $DP=BQ$. Show that
(i) $△APD\cong △CQB$
(ii) $AP=CQ$
(iii) $△AQB\cong △CPD$
(iv) $AQ=CP$
(v) $APCQ$ is a parallelogram

Answer 1

Construction: Join $AC$ to meet $BD$ in $O$.

Therefore, $OB=OD$ and $OA=OC$ ...(1)

(Diagonals of a parallelogram bisect each other)

But $BQ=DP$ ...given
$\therefore OB–BQ=OD–DP$
$\therefore OQ=OP$ ....(2)

Now, in $\square APCQ$,

$OA=OC$ ....from (1)

$OQ=OP$ ....from (2)

$\therefore \square APCQ$ is a parallelogram.

In $△APD$ and $△CQB$,

$AD=CB$ ....opposite sides of a parallelogram

$AP=CQ$ ....opposite sides of a parallelogram
$DP=BQ$ ...given
$△APD\cong △CQB$ ...By SSS test of congruence

$\therefore AP=CQ$ ...c.s.c.t.

$AQ=CP$ ...c.s.c.t. ...(3)

In $△AQB$ and $△CPD$,
$AB=CD$ ....opposite sides of a parallelogram
$AQ=CP$ ...from (3)
$BQ=DP$ ...given

$\therefore △AQB\cong △CPD$ ....By SSS test of congruence