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 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?

[**Remark:** Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into *n* equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide ΔABC into *n* triangles of equal areas.]

#### Solution

Let us draw a line segment AM ⊥ BC.

We know that,

Area of a triangle = 1/2 × Base × Altitude

`"Area "(triangleADE)=1/2xxDExxAM`

`"Area "(triangleABD)=1/2xxBDxxAM`

`"Area "(triangleAEC)=1/2xxECxxAM`

It is given that DE = BD = EC

`⇒ 1/2xxDExxAM=1/2xxBDxxAM=1/2xxECxxAM`

⇒ Area (ΔADE) = Area (ΔABD) = Area (ΔAEC)

It can be observed that *Budhia* has divided her field into 3 equal parts.