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Triangles on the Same Base and between the Same Parallels

iii In the fo...

Question

Question 5 (iii)
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:

(iii) ar(ABC) = 2ar(BEC)

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Solution

(iii)

ar $(Δ A B E) = a r (Δ B E C)$ (Common base BE and BE||AC)
$a r (Δ A B F) + a r (Δ B E F) = a r (Δ B E C)$
We know that,
ar $(Δ B E F) = a r (Δ A F D)$
$a r (Δ A B F) + a r (Δ A F D) = a r (Δ B E C)$
$a r (Δ A B D) = a r (Δ B E C)$
$\frac{1}{2} a r (Δ A B C) = a r (Δ B E C)$
$∴ a r (Δ A B C) = 2 a r (Δ B E C)$

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a r (B D E) = \frac{1}{4} a r (A B C)

a r (B D E) = \frac{1}{4} a r (A B C)

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a r (B D E) = \frac{1}{4} a r (A B C)

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a r (F E D) = \frac{1}{8} a r (A F C)

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