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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.].

Answers (1)
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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(\DeltaBFE) = 2. ar(\DeltaFED) ] 
 \\=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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