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 = 60^o

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 = 60^o and,

∠BDE = 60^o

∠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 (