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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 (iv) ar (BFE) = ar (AFD)

Q: 5        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

(iv)  \small ar(BFE)=ar(AFD)
[Hint: Join EC and AD. Show that  \small BE\parallel AC and  \small DE\parallel AB, etc.]

        

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

\DeltaBDE and \DeltaAED are on the same base ED and between same parallels AB and DE.
Therefore, ar(\DeltaBED) = ar(\DeltaAED)

On subtracting \DeltaEFD from both sides we get

\small ar(BFE)=ar(AFD)

Hence proved.

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