Question

answer this guys(✷‿✷)(✷‿✷)​

Answer 1

$\huge\rm{\bold{\underline{\underline{Anwer:-}}}}$

$\bold{\rm{\underline{\bigstar{Given:-}}}}$

In the given figure-

✳ABCD is a quadrilateral.

✳P, Q, R, S are the points of trisection of the sides AB, BC, CD and DA respectively.

$\bold{\rm{\underline{\bigstar{To\:prove:-}}}}$

PQRS is a parallelogram.

$\bold{\rm{\underline{\bigstar{Proof:-}}}}$

$\large\sf{In\:\triangle ABC -}$

$\large\sf{ \frac{AP}{PB} = \frac{1}{2} ........(i)}$

$\large\sf{and\: \frac{CQ}{QB} = \frac{1}{2 } ........(ii) }$

Since, P and Q are the points of trisection of sides AB and BC respectively.

$\large\sf{From\:equation\:(i) \:and\:(ii) -}$

$\large\sf{\implies\:\frac{AP}{PB} = \frac{CQ}{QB} }$

$\large\sf{Therefore,\: PQ \:ll \:AC}$

(By the converse of Basic Proportionality Theorem. )

$\large\sf{ In\:\triangle ADC -}$

$\large\sf{ \frac{AS}{SD} = \frac{1}{2} ........(iii) }$

$\large\sf{ and\: \frac{CR}{RD} = \frac{1}{2} ........(iv) }$

Since, R and S are the points of trisection of sides CD and DA respectively.

$\large\sf{ From\:equation\:(iii) \:and\:(iv) - }$

$\large\sf{\implies\: \frac{AS}{SD} = \frac{CR}{RD} }$

$\large\sf{ Therefore,\: SR \:ll \:AC. }$

(By the converse of Basic Proportionality Theorem. )

$\large\sf{Therefore, PQ \:ll \:SR. }$

(Since, PQ II AC & SR II AC)

$\large\sf{Similarly,\:by\:joining\:BD\:we \:can\: }$

$\large\sf{ prove \:that \:SP\:II\:RQ. }$

$\large\sf{As \: SR \:ll \:PQ\:and \: SP\:II\:RQ,\:therefore- }$

$\large\sf{ PQRS \:is\: a\: parallelogram. }$

$\large\sf{Hence\:proved. }$