Question

In Fig. 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(ΔBDE) = ar(ΔABC) 

(ii) ar(ΔBDE) = \(\frac{1}{4}\)ar(ΔBAE) 

(iii) ar(ΔBFE) = \(\frac{1}{2}\)ar(ΔAFD) 

(iv) ar(ΔABC) = ar(ΔBEC) 

(v) ar(ΔFED) = \(\frac{1}{8}\)ar(ΔAFC) 

(vi) ar(ΔBFE) = 2ar(ΔEFD)

Answer 1

Given that, 

ABC and BDF are two equilateral triangles 

Let, 

AB = BC = CA = x 

Then, 

BD =\(\frac{x}{2}\) = DE = BF

(i) We have,

(ii) It is given that triangles ABC and BED are equilateral triangles 

∠ACB = ∠DBE = 60°

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,

(iii) Since,

ΔABC and ΔAED are equilateral triangles

Therefore, 

∠ABC = 60° and, 

∠BDE = 60° 

∠ABC = ∠BDE 

AB ‖ DE 

Triangles BED and AED are on the same base ED and between the same parallels AB and DE 

Therefore,

(iv) Since, 

ED is the median of ΔBEC

Therefore,

Area (ΔBEC) = 2 Area (ΔBDE)

Area (ΔBEC) = 2 x \(\frac{1}{4}\) Area (ΔABC)

Area (ΔBEC) = \(\frac{1}{2}\) Area (ΔABC)

Area (ΔABC) = 2 Area (ΔBEC)

(v) Let h be the height of vertex E, 

Corresponding to the side BD on ΔBDE 

Let H be the vertex A, 

Corresponding to the side BC in  ΔABC

From part (i), we have

Area(ΔBDE) = \(\frac{1}{4}\)Area(ΔABC)

\(\frac{1}{2}\) \(\times\) BD \(\times\) h = \(\frac{1}{4}\)(\(\frac{1}{2}\)\(\times\)BC \(\times\)H)

 BD \(\times\) h = \(\frac{1}{4}\)(2 BD \(\times\)H)

h = \(\frac{1}{2}\)H (1)

From part (iii), we have

(vi) Area(ΔAFC)  = Area(ΔAFD) + Area(ΔADC)

= Area(ΔBFE) + \(\frac{1}{2}\)Area(ΔABC)

[Using part (iii) and AD is the median of Area ΔABC]

= Area(ΔBFE) + \(\frac{1}{2}\) \(\times\) 4 Area(ΔBDE)

[Using part (i)]

Area(ΔBFC) = 2 Area(ΔFED) (2)

Area(ΔBDE) = Area(ΔBEF) + Area(ΔFED)

2 Area(ΔFED) + Area(ΔFED)

3 Area(ΔFED) (3)

From above equations,

Area(ΔAFC) = 2 Area(ΔFED) + 2 \(\times\) 3 Area(ΔFED) 

= Area(ΔFED) 

Hence,

Area(ΔFED) = \(\frac{1}{8}\)Area(ΔAFC)