A point E is taken on the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. Prove that `ar(DeltaADF) = ar (ABFC)`.

A point is taken on the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. Prove that : ar(ADF) = ar(ABFC)

M is a point on the side BC of a parallelogram ABCD. DM when produced meets AB produced at N. Prove that (i) (DM)/(MN)=(DC)/(BN)" " (ii) (DN)/(DM)=(AN)/(DC)

P is a point on side BC of a parallelogram ABCD. If DP produced meets AB produced at point L, prove that : DL : DP = AL : DC

E is any point on the side BC of the parallelogram ABCD. AE intersects the diagonal BD at point F. Prove that DF×EF=FB×FA .

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

P is a point on side BC of a parallelogram ABCD. If DP produced meets AB produced at point L, prove that : DP : PL = DC : BL .

In figure, ABCD and AEFD are two parallelograms. Prove that ar (DeltaPEA) = ar (DeltaQFD) .

In figure, ABCD and AEFD are two parallelograms. Prove that ar (DeltaPEA) = ar (DeltaQFD) .

In the figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that ar( Delta AEDF) = ar( Delta ABCDE)

A point D is taken on the side BC of a DeltaABC such that BD = 2DC. Prove that ar(DeltaABD) = 2 ar (DeltaADC)