But how do you get that?

I am very confused: how a single point be the point of trisection of a side of quadrilateral, there must be two points on each side to satisfy the given question.

& if it is point of bisection, then the question become very easy. I think we have solved this type of questions in class 9th.

My best wishes for exams to one and all on the "meritnation".

I can't understand what is meant by that the points are adjacent to A & C. I think this is of no use.

According to me the same is conveyed by the  previous line that the points are on AB, BC, CD, & DA respectively.

If I am wrong,then give the correct explanation.

 hre it is

BP = 2PA

BQ=2QC

DR= 2RC

DS= 2SA

in triangle ADC 

DS/SA = 2PA/SA = 2

and

DR/RC = 2RC/RC = 2

therefre DS/SA = DR/RC

so, SR// AC

in triangle ABC

BP/PA  = 2PA.PA

=2

BQ// QC = 2QC/QC = 2

THEREFORE, BP/PA = BQ/QC

THEREFORE PQ//AC

from the above,we get

SR//PQ

similarly by joinin BD,

we get

QR// PS

therefore, PQRS IS A PARALLELOGRAM. 

thmbz up plz!!

Thanks for the answer.

But sorry, I am unable to give you a thumbs up, may it be some due to my connection problem.

 it is the same question sumukh, its given in rd sharma. the question says tat the given pts are the trisection points. the other line, yes, it is of no use! u r corrct wid dat. but the line is given in the question pronted in the book!

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