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Question 5 (vi)
In the following figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that:
ar(FED)=18ar(AFC)






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Solution


Area (AFC) = area (AFD) + area (ADC)
=ar(BEF)+12ar(ABC) [In (iv), ar(BFE)=ar(AFD); AD is median of ΔABC]
=ar(BFE)+12×4ar(BDE) [In (i), ar(BDE)=14ar(ABC)]
= ar(BEF) + 2ar(BDE)....(5)
Now, by (v),
ar(BFE) = 2ar(FED) …….(6)
ar(BDE) = ar(BFE) + ar(FED)
= 2ar(FED) + ar(FED)
= 3ar(FED) ……(7)
Therefore, from equations (5), (6), and (7), we get:
ar(AFC)=2ar(FED)+2×3ar(FED)=8ar(FED)
ar(AFC)=8ar(FED)
Hence, ar(FED)=18ar(AFC)


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