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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.].

Answers (1)

M manish painkra

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