Concept : 1 Mark
Application : 1 Mark
Proof : 2 Marks
ar(ΔBDE)=ar(ΔAED)
[Since they have common base DE and DE||AB]
Then ar(ΔBDE)−ar(ΔFED)=ar(ΔAED)−ar(ΔFED)
[Subtracting ar (ΔFED) from both sides]
∴ar(ΔBEF)=ar(ΔAFD)....(1)
ar(ΔABD)=ar(ΔABF)+ar(ΔAFD)
ar(ΔABD)=ar(ΔABF)+ar(ΔBEF) [From equation (1)]
ar(ΔABD)=ar(ΔABE)....(2)
AD is the median in ΔABC.
ar(ΔABD)=12ar(ABC)=42ar(ΔBDE)
[Since,ABC and BDE are two equilateral triangles and D is the mid-point of BC]
ar(ΔABD)=2ar(ΔBDE)…….(3)
From (2) and (3),we obtain
2ar(ΔBDE)=ar(ΔABE)
or,ar(ΔBDE)=12ar(ΔABE)