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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 (vi) ar (FED) = 1/8 ar (AFC)

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

(vi) $ar (FED) = \frac{1}{8} ar (AFC)$

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

Views

$ar(\Delta AFC) =$
$\\= ar(\Delta AFD) + ar(\Delta ADC)\\ = ar(\Delta BFE) + 1/2. ar(\Delta ABC)\\ = ar(\Delta BFE) + 1/2\times [4\times ar(\Delta BDE)]\\ = ar(\Delta BFE) + 2\times ar(\Delta BDE)$
$= 2\times ar (\Delta FED) + 2\times [ar(\Delta BFE) + ar(\Delta FED)]$ .....(from part (v) ar($\Delta$BFE) = 2. ar($\Delta$FED) ]
$\\=2ar(\Delta FED) + 2[3\times ar(\Delta FED)]\\ =8\times ar(\Delta FED)$

$ar (FED) = \frac{1}{8} ar (AFC)$
Hence proved.

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