Question

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

Answer 1

Given,

$DP=BQ$

$ABCD$ is parallelogram and $DP=BQ$

In ${△}^{le}APD\&{△}^{le}CQB$

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

$DP=BQ$ ( given )

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

$\therefore$ By SAS congrency,

$△APD\cong △CQB$

$\therefore$ By corresponding parts of congruent triangles.

Hence, proved $AP=CQ.$