Here we have joined " AP " and " CQ " and assume that QP and BD intersect at ' O ' .

Given : P and Q are mid points of ' BC ' and ' AD ' respectively , So

BP = PC =12BC and DQ = QA =12DA and we know BC = DA ( As we know ABCD is a parallelogram and opposite sides are equal to each other )

Then,

BP = PC = DQ = QA --- ( 1 )

i ) We know ABCD is a parallelogram and opposite sides are parallel to each other , So

BC | | DA , So

PC | | QA --- ( 2 ) ( As here PC is part of line BC and QA is a part of line DA and we know BC | | DA )

And

PC = QA ---- ( 3 ) ( From equation 1 )

We know is a pair of opposite sides of any quadrilateral are equal and parallel to each other then that quadrilateral is a parallelogram . Thus from equation 2 and 3 we get :

APCQ is a parallelogram . ( Hence proved )

ii ) We know : BC | | DA , So

BP | | QA --- ( 4 ) ( As here BP is part of line BC and QA is a part of line DA and we know BC | | DA )

And

BP = QA ---- ( 5 ) ( From equation 1 )

We know is a pair of opposite sides of any quadrilateral are equal and parallel to each other then that quadrilateral is a parallelogram . Thus from equation 4 and 5 we get :

ABPQ is a parallelogram .So

AB | | PQ as we know opposite sides are parallel to each other , So

AB | | OQ --- ( 6 ) ( As here OQ is part of line PQ and we know AB | | PQ )

Here we also know ' Q ' is mid point of AD and we consider information " AB | | OQ " from equation 6

Then , In triangle ABD we get from converse of mid point theorem :

' O ' is mid point of BD , So we can say that :

QP is bisect line BD . ( Hence proved )