D and E are points on sides AB and AC respectively of `DeltaABC` such that ar (DBC) = ar (EBC). Prove that `DE II BC`.

D and E are points on sides AB and AC respectively of a DeltaABC such that ar (DeltaBCE) = ar (DeltaBCD) . Show that DE || BC .

D and E are the mid-points of the sides AB and AC respectively of DeltaABC . DE is produced to F. To prove that CF is equal and parallel to DA, we need an additional information which is

Points P, Q and R lie on sides BC, CA and AB respectively of triangle ABC such that PQ || AB and QR || BC, prove that RP II CA.

P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC).

D,E and F are mid-points of the sides BC, CA and AB respectively of triangleABC . Prove that ar(triangleDEF) = 1/4 ar (triangleABC)

D,E and F are mid-points of the sides BC, CA and AB respectively of triangleABC . Prove that ar(||gmBDEF)= 1/2 ar (triangleABC)

In triangleABC , it L and M are points on AB and AC respectively such that LM||BC. Prove that : ar(triangleLCM) = ar(triangleLBM)

In triangleABC , it L and M are points on AB and AC respectively such that LM||BC. Prove that : ar(triangleLBC) = ar(triangleMBC)

If D,E and F are the mid-points of the sides BC,CA and AB, respectively of a DeltaABC and O is any point, show that (i) AD+BE+CF=0

In figure. D and E are the mid-points o sides. AB, AC respectively of triangleABC . If ar(triangleABC) = 256 cm^2 , then find ar(triangleBDE)