In the adjoining figure, BD||CA, E is the midpoint of CA and `BD =(1)/(2) CA`. Prove that
`ar(triangleABC)=2ar(triangleDBC)`.

In the following figure, BD is parallel to CA, E is mid-point of CA and BD = (1)/(2) CA. Prove that : ar.(DeltaABC)=2xxar.(DeltaDBC)

In the following figure, BD is parallel to CA, E is mid-point of CA and BD = (1)/(2) CA. Prove that : ar.(DeltaABC)=2xxar.(DeltaDBC)

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

ABC is a triangle in which D is the midpoint of BC and E is the midpoint of AD. Prove that ar(triangleBED)=(1)/(4)ar(triangleABC) .

The vertex A of triangle ABC is joined to a point D on the side BC. The midpoint of AD is E. Prove that ar(triangleBEC)=(1)/(2)ar(triangleABC) .

D is the midpoint of side BC of triangleABC and E is the midpoint of BD. If O is the midpoint of AE, prove that ar(triangleBOE)=(1)/(8)ar(triangleABC) .

In triangleABC , point D lies on side BC. E is the midpoint of AD. Prove that, ar(EBC)= 1/2 ar(ABC)

In the adjoining figure, DE||BC and AD=1 cm,BD=2 cm.What is the ratio of the area of triangleABC to the area of triangleADE ?

In triangleABC , if D is the midpoint of BC and E is the midpoint of AD then ar(triangleBED) = ?