Question

ABCD is a parallelogram, if p is a point inside the parallelogram then proof that ∆apb+∆pcd=1/2of area of parallelogram ABCD.

Answer 1

first see the diagram
given:a parallelogram ABCD
and p is any point inside it

to prove: ar ∆apb+ ar ∆pcd=1/2 ar of ABCD
construction:join pc ,cd ,pa,and pb
and draw a line mn parallel to ab and cd through p

prooof:now ,
in ∆ pcd and parallelogram dcmn ,
dc is the common base and they lie between same parallels that are
mn parallel to dc
there fore ,
ar ∆pcd =1/2 ar of parallelogram dcmn -(1)( triangl and parallelogram on the same base and between same parellels)

similarly,,∆apb and parallelogram abmn are on same base that is ab and lie between same parellels that are
ab parellel to mn
there fore,
ar.∆apb=1/2 ar parallelogram abmn -(2) ( triangle and a parallelogram on the same base and between same parellels)
now,
add(1) &(2)
ar∆apb+ ar∆pcd=1/2 ar.abmn+ 1/2 ar. cdmn
ar∆apb+ar∆pcd=1/2 ar. parallelogram (abmn+cdmn)
ar∆apb+ar∆pcd 1/2 ar parallelogram ABCD
hence proved
I am sure it will help you friend and if you have any confusion than you can ask in comments
other wise Mark it as brainliest
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Happy Propose Day