ABCD is a parallelogram and O is a point in its interior. Prove that
(i) `ar(triangleAOB)+ar(triangleCOD)`
`=(1)/(2)ar("||gm ABCD").`
(ii) `ar(triangleAOB)+ar(triangleCOD)=ar(triangleAOD)+ar(triangleBOC)`.

Explanation

GIVEN A||gm ABCD and O is a point in its interior.
TO PROVE (i) `ar(triangleAOB)+ar(triangleCOD)=(1)/(2)ar("||gm ABCD"),`
(ii) `ar(triangleAOB)+ar(triangleCOD)=ar(triangleAOD)+ar(triangleBOC)`.
CONSTRUCTION Draw EOF||AB||DC and GOH||AD||BC.
PROOF `triangle`AOB and ||gm EABF being on the same base AB and between the same parallels AB and EF, we have
`ar(triangleAOB)=(1)/(2)ar("||gm EABF")." "...(i)`
Similarly, `ar(triangle COD)=(1)/(2)ar("||gm EFCD")," "...(ii)`
`ar(triangleAOD)=(1)/(2)ar("||gm AHGD")" "...(iii)`
and `ar(triangleBOC)=(1)/(2)ar("||gm BCGH")" "...(iv)`
Adding (i) and (ii), we get
`ar(triangleAOB)+ar(triangleCOD)=(1)/(2)ar("||gm EABF")+(1)/(2)ar("||gm EFCD")`
`=(1)/(2)ar("||gm ABCD")." "...(v)`
Adding (iii) and (iv), we get
`aar(triangleAOD)+ar(triangleBOC)=(1)/(2)ar("||gm AHGD")+(1)/(2)ar("||gm BCGH")`
`=(1)/(2)ar("||gm ABCD")." "...(vi)`
`therefore ar(triangleAOB)+ar(triangleCOD)=ar(triangleAOD)+ar(triangleBOC)`
[from (v) and (vi)].