Question

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

Answer 1

$\Rightarrow$ $ABCD$ is a parallelogram, where $DP=BQ$

Now, $AD\parallel BC$ and $BD$ is transversal

$\Rightarrow$ $\angle ADP=\angle CBQ$ [ Alternate angles ] --- ( 1 )

In $△APD$ and $△CQB,$

$\Rightarrow$ $AD=CB$ [ Opposite sides of parallelogram are equal ]

$\Rightarrow$ $\angle ADP=\angle CBQ$ [ From ( 1 ) ]

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

$\therefore$ $△APD\cong △CQB$ [ By SAS congruence ]