Areas of Parallelograms and Triangles

[agnibho]

Homework Statement

ABC and BDE are two equilateral triangles such that D is the midpoint of BC. If AE intersects BC at F, show that :-
(i) ar(BDE) = 1/4 ar(ABC)
(ii) ar(BDE) = 1/2 ar(BAE)
(iii) ar(ABC) = 2ar(BEC)
(iv) ar(BFE) = ar(AFD)
(v) ar(BFE) = 2ar(FED)
(vi) ar(FED) = 1/2ar(AFC)

Homework Equations

The Attempt at a Solution

There was this hint :- [join EC and AD and then showing BE ll AC and DE ll AB,etc.]

 

[agnibho]

I couldn't upload the diagram.
I tried to follow the hint but from there on I couldn't succeed in solving this sum.
If anyone would tell me how to do this sum I'll be pleased

 

[AC130Nav]

ABC and BDE are two equilateral triangles such that D is the midpoint of BC. If AE intersects BC at F, show that :-
(i) ar(BDE) = 1/4 ar(ABC)
(ii) ar(BDE) = 1/2 ar(BAE)
(iii) ar(ABC) = 2ar(BEC)
(iv) ar(BFE) = ar(AFD)
(v) ar(BFE) = 2ar(FED)
(vi) ar(FED) = 1/2ar(AFC)

There was this hint :- [join EC and AD and then showing BE ll AC and DE ll AB,etc.]

My take on your diagram attached.
To continue, what is the formula for the area of an equilateral triangle?

 

[agnibho]

Area of an equilateral triangle = \sqrt{}3/4*side²

 

[AC130Nav]

Area of an equilateral triangle = \sqrt{}3/4*side²

I assume my diagram was correct?

Anyway, D is at the midpoint of BC, so BD = BC/2
Aabc = \sqrt{}3/4*BC²
Abde = \sqrt{}3/4*BD² = \sqrt{}3/4*(BC/2)²

Finish that and retry the others.

 

[agnibho]

Yes your diagram was correct

 

[agnibho]

Thanks for your help.