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In Figure, A B C\ a n d\ B D E are two ...

In Figure, `A B C\ a n d\ B D E` are two equilateral triangls such that `D` is the mid-point of `B CdotA E` intersects `B C` in `Fdot` Prove that: `a r( B D E)=1/4a r\ (\ A B C)`

Text Solution

Verified by Experts

ABE and BDE are two equilateral triangles.
Considering a as the side of △ABC.
Then,
`ar(ABC)=sqrt(3)/4a^2`
Since D is the midpoint of BC.
So, `BD=a/2`
​ `ar(BDE)=sqrt3/4(a/2)^2=sqrt3/16a^2`
`rArrar(BDE)=1/4ar(ABC)`
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