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#### Question

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

#### Similar questions VIEW ALL

D and E are points on sides AB and AC respectively of ΔABC such that

ar (DBC) = ar (EBC). Prove that DE || BC.

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[Hint : From A and C, draw perpendiculars to BD.]

Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.

ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).

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