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Question

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(BDE)=12ar(BAE)




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Solution

ar(ΔBDE)=ar(ΔAED)
[Since they have common base DE and DE||AB]
Then ar(ΔBDE)ar(ΔFED)=ar(ΔAED)ar(ΔFED)
[Subtracting ar (ΔFED) from both sides]
ar(ΔBEF)=ar(ΔAFD)....(1)

ar(ΔABD)=ar(ΔABF)+ar(ΔAFD)
ar(ΔABD)=ar(ΔABF)+ar(ΔBEF) [From equation (1)]

ar(ΔABD)=ar(ΔABE)....(2)

AD is the median in ΔABC.
ar(ΔABD)=12ar(ABC)=42ar(ΔBDE)
[Since,ABC and BDE are two equilateral triangles and D is the mid-point of BC]

ar(ΔABD)=2ar(ΔBDE).(3)

From (2) and (3),we obtain
2ar(ΔBDE)=ar(ΔABE)
or,ar(ΔBDE)=12ar(ΔABE)



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