2.In Fig. 9.30, D and E are two points on BCsuch that BD DE EC. Show t

1.	2.In Fig. 9.30, D and E are two points on BCsuch that BD DE EC. Show thatar (ABD)-ar (ADE) ar (AEC)Can you now answer the question that you haveleft in the 'Introduction' of this chapter, whetherthe field of Budhia has been actually dividedinto three parts of equal area?Fig. 9.30
Answer» Given: ABC is a Triangle , D & E aretwo Points on BC, Such that BD= DE= EC To Prove: ar (ABD) = ar (ADE) = ar (AEC)Proof: Let AO be the perpendicular to BC. We know that, Area of ∆ =1/2× base × height ar(∆ABD)= ½× BD× AO ar(∆ADE)= ½× DE× AO ar(∆AEC)= ½× EC× AO BD= DE= EC [given] ar(∆ABD)= ar(∆ADE)= ar(∆AEC)

2.In Fig. 9.30, D and E are two points on BCsuch that BD DE EC. Show thatar (ABD)-ar (ADE) ar (AEC)Can you now answer the question that you haveleft in the 'Introduction' of this chapter, whetherthe field of Budhia has been actually dividedinto three parts of equal area?Fig. 9.30

Answer»

Given: ABC is a Triangle , D & E aretwo Points on BC, Such that BD= DE= EC

To Prove:

ar (ABD) = ar (ADE) = ar (AEC)Proof:

Let AO be the perpendicular to BC.

We know that,

Area of ∆ =1/2× base × height

ar(∆ABD)= ½× BD× AO

ar(∆ADE)= ½× DE× AO

ar(∆AEC)= ½× EC× AO

BD= DE= EC [given]

ar(∆ABD)= ar(∆ADE)= ar(∆AEC)