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# In Fig. 9.30, 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? Q.2

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What is the best way for solving the question from class 9th ncert math of Areas of Parallelograms and Triangles chapter of Ncert of exercise 9.4 of math. What is the best way for solving this question please guide me the best way for solving this question. In Fig. 9.30, 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? [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 DABC into n triangles of equal areas.]

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1. Given,

BD = DE = EC

To prove,

ar (△ABD) = ar (△ADE) = ar (△AEC)

Proof,

In (△ABE), AD is median [since, BD = DE, given]

We know that, the median of a triangle divides it into two parts of equal areas

, ar(△ABD) = ar(△AED) —(i)

Similarly,

In (△ADC), AE is median [since, DE = EC, given]