Question

In fig, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that:

(i) ar(BDE)=1/4ar(ABC)

(ii) ar(BDE)=1/2ar(BAE)

(iii) ar(ABC)=2ar(BEC)

(iv) ar(BFE)=ar(AFD)

(v) ar(BFE)=2ar(FED)

(vi) ar(FED)=1/8ar(AFC)

Answer 1

Given : 

ABE and BDE are two equilateral triangles.

Considering a as the side of △ABC.

Then, ar(ABC) = \(\frac{\sqrt{3}}{4}a^2\)

(i)  Since D is the midpoint of BC.  

(ii) Since DE is a median of △BEC and each median divides a triangle in two other  △s of equal area, 

So, ar(BDE) = 1/2 ​ar(BEC) ... (1) 

∠EBC = ∠BCA = 60°

This suggests that they are alterante angles between the lines BE and AC with BC as the transversal.

So, BE∥AC. 

Since △BEC and BAE stand on the same base BE and lie between the same parallel lines BE and AC. 

So, ar(BEC) = ar(BAE)

ar(BDE) = 1/2 ​ar(BEC) (from (1))

ar (ΔBDE) = 1/2 ​ar(BAE). 

(iii) ar(BDE) = 1/2 ar(BEC) 

(ED is a median of △BEC and median divides the triangle in two other △s of equal area. )

ar(BDE) = 1/4 ​ar(ABC)  (Using result from (i))

So, 1/4 ​ar(ABC) = 1/2 ​ar(BEC)

⇒ ar(ABC) = 2ar(BEC)

(iv) ∠ABD = ∠BDE = 60∘ (Angles of equilateral triangle)

These are alternate angles between the lines AB and ED with BD as transversal. 

So, BA || ED. 

So, ar(BDE) = ar(AED) ( Δs  on the same base ED and between the same parallel lines AB and DE) 

Subtracting ar(FED) from both sides,

⇒ ar(BDE) - ar(FED) = ar(AED) - ar(FED)

⇒ ar(BEF) = ar(AFD)

(v) In  Δ  ABC,

By Appolonius theorem,   

AB² + AC² = 2AD² + 2BD²

Since AB = AC,

AD² = AB² - BD²

In  Δ BED,  

By using Appolonius theorem,

Drop perpendicular from EP to BD,

So , EP is the median BED

EP² = DE² - DP²

ar(AFD) = 2ar(EFD) 

Also, ar(BEF) = ar(AFD)

So, ar(BFE) = ar(AFD) = 2ar(EFD)

(vi) ar(BDE) = \(\frac{1}{4}\)​ ar(ABC)