In the given figure, ABCD is a parallelogram P and Q are the mid-point of BC and AD respectively. Prove that
(1). APCQ is a parallelogram
(2). QP bisects BD
Dear Student,
Please find below the solution to the asked query:
We have our diagram , As :
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 = BC and DQ = QA = DA 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 )
Hope this information will clear your doubts about topic.
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Regards
Please find below the solution to the asked query:
We have our diagram , As :
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 = BC and DQ = QA = DA 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 )
Hope this information will clear your doubts about topic.
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Regards