In Fig. 15.86, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. AE intersects BC in F. Prove that
(i) ar(Δ BDE) = 
ar(Δ ABC)
(ii) ar(Δ BDE) = 
ar(Δ BAE)
(iii) ar(Δ BFE) = ar(Δ AFD)
(iv) ar(Δ ABC) = ar(Δ BEC)
(v) ar(Δ FED) = 
ar(Δ AFC)
(vi) ar(Δ BFE) = 2ar(Δ EFD)
Given that,
ABC and BDF are two equilateral triangles
Let,
AB = BC = CA = x
Then,
BD = 
= DE = BF
(i) We have,
Area (
= 
x2
Area (
= 
(
)2
= 
* 
x2
Area (
= 
Area (
(ii) It is given that triangles ABC and BED are equilateral triangles
∠ACB = ∠DBE = 60o
BE ‖ AC (Since, alternate angles are equal)
Triangles BAF and BEC re on the same base BE and between the same parallels BE and AC
Therefore,
Area (
= Area (
Area (
= 2 Area (
(Therefore, ED is the median)
Area (
= 
Area (
BAE)
(iii) Since,
are equilateral triangles
Therefore,
∠ABC = 60o and,
∠BDE = 60o
∠ABC = ∠BDE
AB ‖ DE
Triangles BED and AED are on the same base ED and between the same parallels AB and DE
Therefore,
Area (
= Area (
Area (
- Area (
= Area (
- Area (
Area (
= Area (
(iv) Since,
ED is the median of 
Therefore,
Area (
= 2 Area (
Area (
= 2 * 
Area (
Area (
Area (
Area (
= 2 Area (
(v) Let h be the height of vertex E, corresponding to the side BD on 
Let H be the vertex A, corresponding to the side BC in 
From part (i), we have
Area (
= 
Area (
* BD * h = 
(
* BC * H)
= 
(2 BD * H)
h = 
H (1)
From part (iii), we have
Area (
= Area (
= 
* PD * H
= 
* FD * 2h
= 2 (
* FD * h)
= 2 Area (
(vi) Area (
= Area (
+ Area (
= Area (
+ 
Area (
[Using part (iii) and AD is the median of Area 
= Area (
+ 
* 4 Area (
[Using part (i)]
Area (
= 2 Area (
(2)
Area (
= Area (
+ Area (
2 Area (
+ Area (
3 Area (
(3)
From above equations,
Area (
= 2 Area (
+ 2 * 3 Area (
)
= Area (
)
Hence,
Area (
= 
Area (