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
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
Using (2), (3) and (4), we get
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)
:.by (SAS rule) tri DPQ=tri APQ
2) now join DB . Then we get two equal triangles triDBC andtriABD
and we know PQ is 1/d is1/3 of BDC...2
from 1and 2 we get area triangle pod=1/6area of ABCD