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Triangles on the Same Base and between the Same Parallels

iv In the fol...

Question

Question 5 (iv)
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:

(iv) ar(BFE) = ar(AFD)

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Solution

(iv) It is seen that $Δ B D E$ and $Δ A E D$ lie on the same base (DE) and between the
parallels DE and AB.
$\Rightarrow a r (Δ B D E) = a r (Δ A E D)$
Then $a r (Δ B D E) - a r (Δ F E D) = a r (Δ A E D) - a r (Δ F E D)$ [Subtracting $a r (Δ F E D)$ from both sides]
$∴ a r (Δ B F E) = a r (Δ A F D)$

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Similar questions

A B C

B D E

D

B C

A E

B C

F

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

a r (B D E) = \frac{1}{2} a r (B A E)

a r (A B C) = 2 a r (B E C)

a r (B F E) = a r (A F D)

a r (B F E) = 2 a r (F E D)

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

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

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

a r (B D E) = \frac{1}{2} a r (B A E)

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