In Fig, $D$ and $E$ are two points on $B C$ such that $B D = D E = E C$ . Show that $a r (A B D) = a r (A D E) = a r (A E C)$ .
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?

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Solution

Consider $A P$ as the perpendicular to $B C$ .
$A P$ is the height of all three triangles $A B D, A D E$ and $A E C$ .
$a r (A B D) = \frac{1}{2} \times B D \times A P$
$a r (A D E) = \frac{1}{2} \times D E \times A P$
and $a r (A E C) = \frac{1}{2} \times E C \times A P$
Since $B D = D E = E C$
$a r (A B D) = a r (A D E) = a r (A E C)$

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In Fig, 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?

Consider AP as the perpendicular to BC. AP is the height of all three triangles ABD,ADE and AEC. ar(ABD)=12×BD×APar(ADE)=12×DE×APand ar(AEC)=12×EC×APSince BD=DE=ECar(ABD)=ar(ADE)=ar(AEC)