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Congruency Criterion for Triangles AAS, ASA, SSS, SAS

Question 5 iv...

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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Q. 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)

Q. Question 5 (i)
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

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

Q. Question 5 (ii)
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:

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

Q. 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)

Q. Question 5 (vi)
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:

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

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