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  • In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that (ii) ar (BDE) = 1/2 ar (BAE)

Q: 5    In Fig. \small 9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that

(ii)  \small ar(BDE)=\frac{1}{2}ar(BAE)

[Hint: Join EC and AD. Show that \small BE\parallel AC and  \small DE\parallel AB, etc.]

        

Answers (1)

best_answer


Since \DeltaABC and \Delta BDE are equilateral triangles.
Therefore, \angleACB = \angleDBE = 60^0
\Rightarrow BE || AC

\DeltaBAE and \DeltaBEC are on the same base BE and between same parallels BE and AC.
Therefore, ar (\DeltaBAE) = ar(\DeltaBEC)
 \Rightarrow ar(\DeltaBAE) = 2 ar(\DeltaBED)   [since D is the meian of \DeltaBEC ]
\Rightarrow \small ar(BDE)=\frac{1}{2}ar(BAE)
Hence proved.

Posted by

manish painkra

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