In the figure P, Q, R and S are points on the sides of quadrilateral ABCD such that A B D C O 1 2 A D C E B F these points divide the sides AB, CB, CD and AD in the ratio 2:1. Prove that PQRS is a parallelogram.
Answers
Answer:
Given A quadrilateral ABCD in which P,Q,R and S are the points of trisection of sides AB,BC,CD and DA respectively and are adjacent to A and C.
To prove PQRS is a parallelogram i.e., PQ||SR and QR||PS.
Construction Join AC.
Proof Since P,Q,R and S are the points of trisection of AB,BC,CD and DA respectively.
∴ BP=2PA,BQ=2QC,DR=2RC and, DS=2SA
In △ADC, we have
SA
DS
=
SA
2SA
=2 and,
RC
DR
=
RC
2RC
=2
⇒
SA
DS
=
RC
DR
⇒ S and R divide the sides DA and DC respectively in the same ratio.
⇒ SR∣∣AC {By the converse of Thale's Theorem]
In △ABC, we have
PA
BP
=
PA
2PA
=2 and
QC
BQ
=
QC
2QC
=2
⇒
PA
BP
=
QC
BQ
⇒ P and Q divide the sides BA and BC respectively in the same ratio.
⇒ PQ∣∣AC ........(ii)
From equations (i) and (ii), we have
SR∣∣AC amd PQ∣∣AC
⇒ SR∣∣PQ
Similarly, by joining BD, we can prove that
⇒ QR∣∣PS
Hence, PQRS is a parallelogram
Answer:
Join AC
In △ABC, we have
⇒
PB
AP
=
2
1
⇒ And
QB
CQ
=
2
1
⇒
AB
AP
=
QB
CQ
∴ By using converse of basic proportionality theorem, we have
PQ∥AC ------ ( 1 )
In △ADC, we have
⇒
AS
DS
=
1
2
And
RC
DR
=
1
2
⇒
AS
DS
=
RC
DR
∴ By using converse of basic proportionality theorem, we have
SR∥AC ------ ( 2 )
Comparing ( 1 ) and ( 2 ), we get
⇒ SR∥PQ
Similarly, we can prove
⇒ SP∥RQ
∴ PQRS