In the adjoining figure, E is a point on the median `AD` of a `DeltaABC.` Show that ar `(DeltaABE)=(DeltaACE).`

In the figure 11.23, E is any point on median AD of a Delta ABC. Show that ar (ABE) = ar (ACE).

In Fig.9.23,E is any point on median AD of a Delta ABC Show that ar (ABE)=ar(ACE)

In the adjoining figure, P is the point in the interior of a parallelogram ABCD.Show that ar(DeltaAPB) ar(DeltaPCD) =1/2ar(||gm ABCD)

In the Given figure, E is any point on median AD of a triangle ABC Show that ar (ABE) = ar (ACE)

In the figure, E is any point on median AD of a triangleABC . Show that ar( triangleABE )= ar( triangleACE ).

E is any point on median AD of a triangle ABC . Show that ar (ABE) = ar (ACE).

E is any point on median AD of DeltaABC . If ar (DeltaABE) = 10 cm^2 then ar (DeltaACE) is :

E is any point on median AD of DeltaABC . If ar (DeltaABE) = 10 cm^2 then ar (DeltaACE) is :

If E be the mid-point of the median AD of DeltaABC , then prove that DeltaBED=1/4DeltaABC .

In the figure D, E are points on the sides AB and AC respectively of DeltaABC such that ar(DeltaDBC) = ar(DeltaEBC) . Prove that DE || BC.