Question

ABCD is a quad. And P, Q, R, S are the points of trisection of the sides AB, BC, CD and DA respectively and are adjacent to A and C. Prove that PQRS is a parallelogram. ​

Answer 1

$\huge \underline{ \underline \mathfrak{solution - }}$

Given - a quad. ABCD in which P, Q, R, S are the points of trisection of AB, BC, CD and DA respectively.

To prove - PQRS is a parallelogram.

In the given figure, it is already constructed i.e. AC is joint.

Proof -

$\huge \mathfrak{in \triangle \: BAC }$$\huge \mathfrak { \frac{BP}{PA} } = \frac{BQ}{QC } \: = \frac{2}{1 } \\ \therefore \: PQ \parallel \: AC - - - (1) \: (by \: converse \: of \: thales \: theorem)$

$\huge \mathfrak{also \: in \triangle \: DAC }$

$\huge{ \frac{DS }{SA } = \frac{DR }{AC } = \frac{2}{1} } \\ \therefore \:SR \parallel \: AC - - - -(2) \: \: (by \: converse \: of \: thales \: theorem)$

Thus, PQ||SR. [from (1)&(2)]

Similarly, by joining DB we can prove that SP ||RQ.

Hence PQRS is a parallelogram.

$\huge \underline{ \underline \mathfrak{proved}}$

Answer 2

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

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

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)

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

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

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

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

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