Question

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
i) ar (BDE)=ar (ABC)
(ii) ar (BDE)=ar (BAE)
(iii) ar (ABC)=2 ar (BEC)
(iv) ar (BFE) = ar (AFD)
(v) ar (BFE) = 2 ar (FED)
(vi) ar (FED)=1/8ar (AFC)
[Hint: Join EC and AD. Show that BE|| AC and DE | AB, etc.)​

Answer 1

Answer:

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