In Fig. 30 D and E are to points on BC such that `BD = DE = EC.` Show that `ar(ABD)= ar(ADE) = ar(AEC).`

In the following Figure, D and E are two points on BC such that BD = DE = EC . Show that ar (ABD) = ar (ADE) = ar(AEC) can you now answer the equation that you left in the "Introduction " of the capther, whether the filed of Budhia has been actually divided into three parts of equal area ?

In, D and E are two points on BC such that BD = De = EC. Show that ar (ABD) = ar (ADE) = ar (AEC). Can you now answer the question that you have left in the 'introduction' of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?

E is any point on median AD of a triangle 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)

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

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

AD is one of the medians of a ABC*X is any point on AD. Show that ar (ABX)=ar(ACX)

In the figure, ABCD is a quadrilateral. AC is the diagonal and DE || AC and also DE meets BC produced at E. Show that ar(ABCD) = ar (DeltaABE).

In the figure, ABCD is a quadrilateral. AC is the diagonal and DE || AC and also DE meets BC produced at E. Show that ar(ABCD) = ar (DeltaABE).

In the figure, ABCD is a quadrilateral. AC is the diagonal and DE || AC and also DE meets BC produced at E. Show that ar(ABCD) = ar (DeltaABE).