ABCD is a parallelogram. O is any point on AC. PQ is parallel to AB and LM is parallel to AD. Prove that ar(llgmDLOP) = ar (llgm BMOQ)
Attachments:
Answers
Answered by
16
Since,a diagonal of a parallelogram divides it into two Triangles of equal area.
Since, AO & OC are diagonals of a parallelogram AMOP & OQCL
ar(∆APO) = ar(∆AMO) ………..(1)
ar(∆OLC) = ar(∆OQC) ………..(2)
ar(∆ADC) = ar(∆ABC)
ar(∆APO) + ar(||gmDLOP) +ar(∆OLC) =
ar(∆AMO) + ar(||gm BMOP) +ar(∆OQC) …(3)
Subtract eq 1 & 2 from 3
ar(||gmDLOP) = ar(||gm BMOQ)
Please please mark this answer as brainiest answer please
Since, AO & OC are diagonals of a parallelogram AMOP & OQCL
ar(∆APO) = ar(∆AMO) ………..(1)
ar(∆OLC) = ar(∆OQC) ………..(2)
ar(∆ADC) = ar(∆ABC)
ar(∆APO) + ar(||gmDLOP) +ar(∆OLC) =
ar(∆AMO) + ar(||gm BMOP) +ar(∆OQC) …(3)
Subtract eq 1 & 2 from 3
ar(||gmDLOP) = ar(||gm BMOQ)
Please please mark this answer as brainiest answer please
pinkijyotthika:
Please mark this answer as brainiest answer please
Similar questions