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

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Solution

$a r (Δ B D E) = a r (Δ A E D)$
[Since they have common base DE and DE||AB]
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 ar $(Δ F E D)$ from both sides]
$∴ a r (Δ B E F) = a r (Δ A F D) . . . . (1)$

$a r (Δ A B D) = a r (Δ A B F) + a r (Δ A F D)$
$a r (Δ A B D) = a r (Δ A B F) + a r (Δ B E F)$ [From equation (1)]

$a r (Δ A B D) = a r (Δ A B E) . . . . (2)$

AD is the median in $Δ A B C$ .
$a r (Δ A B D) = \frac{1}{2} a r (A B C) = \frac{4}{2} a r (Δ B D E)$
[Since,ABC and BDE are two equilateral triangles and D is the mid-point of BC]

$a r (Δ A B D) = 2 a r (Δ B D E) \dots \dots . (3)$

From (2) and (3),we obtain
$2 a r (Δ B D E) = a r (Δ A B E)$
$o r, a r (Δ B D E) = \frac{1}{2} a r (Δ A B E)$

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