Question 5 (iii)
D, E and F are respectively the mid-points of the sides BC, CA and AB of triangle ABC Show that:
$(i i i) a r e a (B D E F) = \frac{1}{2} a r e a (A B C)$

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Solution

F ,E are the midpoints of the sides AB and AC respectively.
$∴ F E | | B C a n d F E = \frac{1}{2} B C$
$\Rightarrow F E | | B C a n d F E = B D [D i s m i d p o i n t o f B C] . . . . . (1)$
Similarly D, E are the midpoints of BC and AC respectively.
$∴ E D | | F B a n d E D = \frac{1}{2} A B$
$\Rightarrow E D | | F B a n d E D = F B [F i s m i d p o i n t o f A B] . . . . . (2)$
From (1) and (2) BDEF is a parallelogram.
We know that,
area(ABC) = 4 area(DEF)
$= 4 a r e a (\frac{1}{2} a r e a B D E F)$ [For a parallelogram BDEF, diagonal FD divides it into two triangles of equal area]
= 2 area(BDEF)
i.e. area $(B D E F) = \frac{1}{2} a r e a (A B C)$

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