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

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Solution

Given : In $△ A B C,$ D is a point on BC such that BD = 2 DC
AD is joined
To prove : ar $(△ A B D) = 2 a r (△ A D C)$
Construction : Draw $A L ⊥ B C$

Proof : area $(△ A B D) = \frac{1}{2} \times b a s e \times h e i g h t = \frac{1}{2} B D \times A L$
and area $(△ A D C) = \frac{1}{2} D C \times A L$
But $B D = 2 D C \Rightarrow D C = \frac{1}{2} B D ∴ a r e a (△ A D C) = \frac{1}{2} (\frac{1}{2} B D) \times A L = \frac{1}{2} [\frac{1}{2} B D \times A L] = \frac{1}{2} [a r (△ A B D)] ∴ a r (△ A D C) = 2 a r e a (△ A B D)$

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

A point D is taken on the side of BC of ∆ABC such that BD = 2DC. Then

D

E

B C

△ A B C

B D = C E

A D = A E

△ A B D ≅ △ A C E

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B

D

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(B D > B C),

B D = 2 D C

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