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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.]

Views

Since $\Delta$ABC and $\Delta$ BDE are equilateral triangles.
Therefore, $\angle$ACB = $\angle$DBE = $60^0$
$\Rightarrow$ BE || AC

$\Delta$BDE and $\Delta$AED are on the same base ED and between same parallels AB and DE.
Therefore, ar($\Delta$BED) = ar($\Delta$AED)

On subtracting $\Delta$EFD from both sides we get

$\small ar(BFE)=ar(AFD)$

Hence proved.

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