In the figure, ABC adn BDC are two equilateral triangles such that D is the mid point of BC. Ae intersects BC in F. (i)ar(△BDE)=14ar(△ABC)(ii)ar(△BDE)=12ar(△BAE)(iii)ar(△BFE)=ar(△AFD)(iv)ar(△ABC)=2ar(△BEC)(v)ar(△BFE)=2ar(EFD)(vi)ar(△FED)=18ar(triangleAFC)

In the figure, ABC adn BDC are two equilateral triangles such that D i

1.	In the figure, ABC adn BDC are two equilateral triangles such that D is the mid point of BC. Ae intersects BC in F. (i)ar(△BDE)=14ar(△ABC)(ii)ar(△BDE)=12ar(△BAE)(iii)ar(△BFE)=ar(△AFD)(iv)ar(△ABC)=2ar(△BEC)(v)ar(△BFE)=2ar(EFD)(vi)ar(△FED)=18ar(triangleAFC)
Answer» In the figure, ABC adn BDC are two equilateral triangles such that D is the mid point of BC. Ae intersects BC in F. $(i) a r (△ B D E) = \frac{1}{4} a r (△ A B C) (i i) a r (△ B D E) = \frac{1}{2} a r (△ B A E) (i i i) a r (△ B F E) = a r (△ A F D) (i v) a r (△ A B C) = 2 a r (△ B E C) (v) a r (△ B F E) = 2 a r (E F D) (v i) a r (△ F E D) = \frac{1}{8} a r (t r i a n g l e A F C)$