In parallelogram $$ABCD$$, two points $$P$$ and $$Q$$ are taken on diagonal $$BD$$ such that
$$DP=BQ$$
proof:(i)In $$\triangle APD$$ and $$\triangle {CQB}$$,
$$\therefore AD\parallel BC$$
opposite sides of parallelogram $$ABCD$$ and a transversal $$BD$$ intersect them
$$\therefore \angle{ ABD}=\angle{ CBD}$$ alternate interior angles
$$\Rightarrow \angle ADP=\angle CBQ$$ ....$$(1)$$
$$ DP=BC$$ ....$$(2)$$
$$ AD=CB$$......$$(3)$$
opposite sides of parallelogram $$ABCD$$
From $$\left(1\right) , \left(2\right)$$ and $$\left(3\right)$$
$$\triangle APD\cong\triangle CQB$$ by SAS congruence
(ii)$$\therefore \triangle APD =\triangle CQB$$ proved in $$(1)$$ above
$$\therefore AP=CQ$$
(iii)In $$\triangle AQB$$ and $$\triangle CPD$$
$$\therefore AB||CD$$ since opposite sides of parallelogram $$ABCD$$ and a transversal $$BD$$ intersects them
$$\therefore \angle ABD=\angle CDB$$ as alternate interior angles
$$\Rightarrow \angle ABQ=\angle CDP$$
$$\Rightarrow QB=PD$$ (given)
$$\Rightarrow AB=CD$$ opposite sides of parallelogram $$ABCD$$
$$\therefore \triangle {AQB}\cong\triangle{ CPD}$$ by SAS congruence rule
$$(iv)\therefore \triangle AQB=\triangle CPD$$ proved in $$\left(iii\right)$$ above
$$\therefore AQ=CP $$ by CPCT
$$(v)\therefore$$ the diagonals of a parallelogram bisects each other
$$\therefore OB=OD$$
$$\therefore OB-BQ=OD-DP$$
$$\therefore BQ=DP$$ (given)
$$\therefore OQ=OP.....(1)$$
Also, $$OA=OC.......(2) $$
$$\therefore$$ diagonals of a parallelogram bisects each other
From $$(1)$$ and $$(2), APCQ$$ is a parallelogram