Question

In the given figure, 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) = $\frac{1}{4}$ ar (Δ ABC)

(ii) ar (Δ BDE) = $\frac{1}{2}$ ar (Δ BAE)

(iii) ar (Δ BFE) =ar (Δ AFD)

(iv) ar (ΔABC) = 2 ar (Δ BEC)

(v) ar (Δ FED) = $\frac{1}{8}$ ar (Δ AFC)

(vi) ar (Δ BFE) = 2 ar (EFD)

Answer 1

Given:

(a) ΔABC and Δ BDE are two equilateral triangles

(b) D is the midpoint of BC

(c) AE intersect BC in F.

To prove:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Proof: Let AB = BC = CA = x cm.

Then BD = = DE = BE

(i) We have

(ii) We Know that ΔABC and ΔBED are equilateral triangles

BE||AC

(iii) We Know that ΔABC and ΔBED are equilateral triangles

AB || DE

(iv) Since ED is a median of Δ BEC

(v) We basically want to find out FD. Let FD = y

Since triangle BED and triangle DEA are on the same base and between same parallels ED and BE respectively. So

Since altitude of altitude of any equilateral triangle having side x is

So

…… (1)

Now

…… (2)

From (1) and (2) we get

(vi) Now we know y in terms of x. So

……. (3)

…… (4)

From (3) and (4) we get