The base BC of triangle ABC is divided at D so that BD = `(1)/(2)` DC.
Prove that area of `DeltaABD=(1)/(3)` of the area of `DeltaABC`.`

In triangle ABC, D is a point in side BC such that 2BD=3DC . Prove that the area of triangle ABD = (3)/(5)xx "Area of" DeltaABC

Any point D is taken on the side BC of a Delta ABC and AD is produced to E such that AD = DE , prove that area of Delta BCE = area of Delta ABC ,

In the adjoining figure, BD = DC and AE = ED. Prove that area of DeltaACE = (1)/(4) " area of " DeltaABC

The medians of Delta ABC intersect at point G. Prove that: area of Delta AGB = (1)/(3) xx " area of " Delta ABC

In the adjoining figure, D is the mid-point of side AB of Delta ABC and P be any point on side BC. If CQ ||PD , then prove that: area of Delta BPQ = (1)/(2) xx " area of " Delta ABC

D is the mid-point of side AB of the triangle ABC.E is the mid-point of CD and F is the mid-point of AE. Prove that 8 xx " area of " (DeltaAFD) = " area of " Delta ABC

D, E and F are the mid-points of the sides BC, CA and AB respectively of triangle ABC. Prove that: area of BDEF is half the area of Delta ABC.

The side BC of triangles ABC is produced to D so that CD = AC . If the angle BAD = 109 ^(@) and the ACB = 72^(@) prove that BD is greater than AD.

In a triangle ABC,D is the mid - point of BC, AD is produced upto E so that DE = AD. Prove that : AB=EC

In the following figure G, is centroid of the triangles ABC. Prove that Area (Delta " AGB " )= (1)/(3) xx " Area " (Delta ABC)