Question

In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP$=$BQ. Show that $\Delta$AQB$\cong \Delta$CPD.

Answer 1

Here, $ABCD$ is a parallelogram.

$AB\parallel DC$ and $BD$ is a transversal.

$\therefore$ $\angle ABQ=\angle CDP$ [ Alternate angles ] ---- ( 1 )

In $△AQB$ and $△CPD,$

$\Rightarrow$ $AB=CD$ [ Opposite sides of parallelogram are equal ]

$\Rightarrow$ $\angle ABQ=\angle CDP$ [ From ( 1 ) ]

$\Rightarrow$ $BQ=DP$ [ Given ]

$\therefore$ $△AQB\cong △CPD$ [ By SAS congruence ]