In the adjoining figure, ABCD is a parallelogram and P is any points on BC. Prove that
`ar(triangleABP)+ar(triangleDPC)=ar(trianglePDA)`.

In the figure. ABCD is a parallelogram and O is any point on BC . Prove that ar (triangleABO) + ar (triangleDOC) = ar(triangleODA)

In the adjoining figure, ABCD is a parallelogram. Points P and Q on BC trisect BC. Prove that ar(triangleAPQ)=ar(triangleDPQ)=(1)/(6)ar(triangleABCD) .

In the given figure, ABCD is a parallelogram. E and F are any two points on AB and BC, respectively. Prove that ar (triangleADF)=ar(triangleDCE) .

In the adjoining figure, MNPQ and ABPQ are parallelogram and T is any point on the side BP. Prove that (i) ar(MNPQ)=ar(ABPQ) (ii) ar(triangleATQ)=(1)/(2)ar(MNPQ) .

ABCD is a parallelogram. P is any point on CD. If ar(triangleDPA) = 15 cm^2 and ar(triangleAPC) = 20 cm^2 , the ar (triangleAPB) =

ABCD is a parallelogram. X and Y are mid-points of BC and CD. Prove that ar(triangleAXY)=3/8ar(||^[gm] ABCD)

ABCD is a parallelogram and P is a point of bar(CD) . Show that ar(/_\ABP) = ar(/_\ADP+ /_\BCP) .

In the figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that ar(PQRS) = ar(ABRS)

ABCD is a parallelogram and a line through A meets DC at P and BC (produced) at Q. Prove that ar(triangleBPC) = ar(triangleDPQ)

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) .