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Congruent Figures

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Question 2
In a triangle ABC, E is the mid-point of the median AD.
Show that ar $(Δ B E D) = \frac{1}{4} a r (Δ A B C)$

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Solution

ED is the median of $Δ E B C$ .
$∴ a r e a (Δ B E D) = a r e a (Δ E C D) . . . . . (1)$
BE is the median of $Δ A B D$ .
$∴ a r e a (Δ B E D) = a r e a (B E A) \dots \dots . (2)$
CE is the median for $Δ A D C$ .
$a r e a (Δ E C D) = a r e a (Δ E A C) \dots \dots (3)$
i,e $a r e a (Δ B E D) = a r e a (Δ E A C)$ [ From (1)]……(4)
Now, area $(Δ A B C) = a r e a (B E A) + a r e a (Δ B E D) + a r e a (Δ E C D) + a r e a (Δ E A C)$
$a r e a (Δ A B C) = a r e a (Δ B E D) + a r e a (Δ B E D) + a r e a (Δ B E D) + a r e a (Δ B E D) [f r o m (1), (2), (3) a n d (4)]$
$a r e a (A B C) = 4 \times a r e a (Δ B E D)$
$\Rightarrow a r e a (Δ B E D) = \frac{1}{4} a r (Δ A B C)$

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Congruent Figures

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