CBSE - Mathematics - Areas of Parallelograms and Triangles

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) = 2 ar(BEC)

Area (ΔBDE) = Area (ΔAED) (Common base DE and DE||AB)

Area (ΔBDE) - Area (ΔFED) = Area (ΔAED) - Area (ΔFED)

Area (ΔBEF) = Area (ΔAFD) (1)

Area (ΔABD) = Area (ΔABF) + Area (ΔAFD)

Area (ΔABD) = Area (ΔABF) + Area (ΔBEF) [From equation (1)]

Area (ΔABD) = Area (ΔABE) (2)

AD is the median in ΔABC.

From (2) and (3), we obtain
2 ar (ΔBDE) = ar (ΔABE)

or ar(BDE) = 1/2 ar(BAE)