The base BC of ABC is divided at D such that BD-1-2DC-Prove that ar-ABD-1-3ar-ABC-

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Triangles on the Same Base and between the Same Parallels

The base BC o...

Question

The base BC of $△ A B C$ is divided at D such that BD= $\frac{1}{2} D C .$ .Prove that $a r (△ A B D) = \frac{1}{3} a r (△ A B C)$ .

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B C

A B C

D

B D = \frac{1}{2} D C

△ A B D = \frac{1}{3}

△ A B C

A point D is taken on the side BC of a $△ A B C$ such that BD = 2DC. Prove that ar $(△ A B D) = 2 a r (△ A D C)$

Δ A B C

B D = \frac{1}{2} B C

(Δ A B D) = \frac{1}{2} a r (Δ A D C)

\frac{1}{2}

Δ A B D = \frac{1}{3}

Δ A B C

B D = \frac{1}{2} D C

ar (∆ A B D) = \frac{1}{3} \times ar (∆ A B C)

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