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(N/A) Let us draw $AF$ perpendicular to $BC$ such that $AF$ is the height of $\Delta ABD$,$\Delta ADE$,and $\Delta AEC$.We know that the area of a tri

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(N/A) Let us draw $AF$ perpendicular to $BC$ such that $AF$ is the height of $\Delta ABD$,$\Delta ADE$,and $\Delta AEC$.
We know that the area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
Therefore,$ar(ABD) = \frac{1}{2} \times BD \times AF$.
Similarly,$ar(ADE) = \frac{1}{2} \times DE \times AF$.
And $ar(AEC) = \frac{1}{2} \times EC \times AF$.
Since it is given that $BD = DE = EC$,we can substitute these values.
Thus,$\frac{1}{2} \times BD \times AF = \frac{1}{2} \times DE \times AF = \frac{1}{2} \times EC \times AF$.
This implies $ar(ABD) = ar(ADE) = ar(AEC)$.
Yes,since the altitudes (heights) of all the triangles are the same and their bases are equal,the areas are equal. Therefore,Budhia can use this result to divide her land into three equal parts.

In the 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 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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In the 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 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?