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The base BC of △ ABC is divided at D such that BD = ½ DC. Prove that ar (△ ABD) = 1/3 × ar (△ ABC).

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Construct AE ⊥ BC

We know that

Area of △ ABD = ½ × BD × AE ……. (1)

Area of △ ABC = ½ × BC × AE …….. (2)

It is given that

BD = ½ BC

We get

BC = BC + DC

It can be written as

BC = BD + 2BD

So we get

BC = 3BD

By division

BD = 1/3 BC …….. (3)

Using equation (1) we get

Area of △ ABD = ½ × BD × AE

By substituting (3)

Area of △ ABD = ½ × 1/3 × BC × AE

So we get

Area of △ ABD = 1/3 (1/2 × BC × AE)

Substituting equation (2)

Area of △ ABD = 1/3 × Area of △ ABC

Therefore, it is proved that ar (△ ABD) = 1/3 × ar (△ ABC).

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