Question

P is any point on side AD of a parallelogram ABCD(fig12.34)
Prove that:ar(∆PAB) + ar(∆PCD)= ar(∆PBC)

Answer 1

Hello Mate!

Since ||gm ABCD and triangle PBC lie on same base so

ar( tri PBC ) = 1/2 ar( ||gm ABCD )

2 ar( tri PBC ) = ar( ||gm ABCD )

ar( ||gm ABCD ) = ar( tri PCD ) + ar( tri PAB ) + ar( tri PBC )

2 ar( tri PBC ) = ar( tri PCD ) + ar( tri PAB ) + ar( tri PBC )

2ar( tri PBC ) - ar( tri PBC ) = ar( tri PCD ) + ar( tri PAB )

ar( tri PBC ) = ar( tri PCD ) + ar( PAB )

Hence Proved ( Q.E.D )

Sorry for complicated abbriviations.

Tri stands for triangle, ar for area and ||gm for parallelogram.

Have great future ahead!