ABCD is a parallelogram.P and Q on BC trisects it.Prove that ar(APQ) =ar(DPQ)=1/6ar(ABCD)

Given, ABCD is a parallelogram in which points P and Q trisects the BC.

⇒ BP = PQ = QC

To prove:

Construction: Through P and Q, draw PR and QS parallel to AB and CD.


Now, ΔAPD and ΔAQD lie on the same base AD and between the same parallel AD and BC.

∴ ar (ΔAPD) = ar (ΔAQD)

⇒ ar (ΔAPD) - ar (ΔAOD) = ar (ΔAQD) - ar (ΔAOD)  [on subtracting ar (ΔAOD) from both sides]

 ⇒ ar (ΔAPO) =  ar (ΔOQD)  ..... (1)

 ⇒ ar (ΔAPO) + ar (ΔOPQ) =  ar (ΔOQD) + ar (ΔOPQ)  [on adding ar (ΔOPQ) on both sides]

 ⇒ ar (ΔAPQ) = ar (ΔDPQ)  ..... (2)

Again, ΔAPQ and parallelogram PQSR are on the same base PQ and between same parallels PQ and AD

 .... (3)


Using (2), (3) and (4), we get

[Hence proved]

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1)  In this parallelogram

DQ=AP(opposite sides of parallelogram equal and parallel and BP and CQ are 1/3 of BC)

similarly ( DP=AQ)

angPAQ=angPDQ (SAS rule) tri DPQ=tri APQ

2) now join DB . Then we get  two equal triangles triDBC  andtriABD

meanse triDBC=1/2abcd...1

and we know PQ is 1/d is1/3 of BDC...2

from 1and 2 we get area triangle pod=1/6area of ABCD

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