In figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC If AE - Maths - Areas of Parallelograms and Triangles

Question

In figure, ABC and BDE are two equilateral triangles such that D
is the mid-point of BC If AE intersects BC at F, prove that
ar(ABE)=1/2ar(ABC)

Manbar Singh · Accepted Answer

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Since ABC is an equilateral ∆, thenAB = BC = CA
Let AB = BC = CA = x
Now, ar∆ABC = 34AB2 = 34x2Now, D is the mid-point of BC
ThenBD = BC2 = x2Since BDE is an equilateral ∆, thenBD = DE = BE = x2ar∆BDE = 34BD2 = 34x22 = 316x2Now,ar∆ABCar∆BDE = 34x2316x2 = 4⇒ar∆ABC = 4ar∆BDE   
1

Now, ABC is an equilateral ∆, so ∠ACB = 60°Also, BDE is an equilateral ∆, so ∠DBE = 60°Now, ∠ACB = ∠DBEBut ∠ACB and ∠DBE are alternate interior angles made by the transversal BC with linesAC and BE and are equal
So, BE ∥ AC
Now, ∆ABE and ∆BCE have same base BE and between same parallel lines BE and AC, thenar∆ABE = ar∆BCE    
2Since ED is median to side BC of ∆BCE, thenar∆BDE = 12 ar∆BCE     Median of a ∆ divides it into 2 ∆'s of equal areas⇒ar∆BCE = 2 ar∆BDENow, from 2, we getar∆ABE =2 ar∆BDE⇒ ar∆BDE = 12ar∆ABE     
3Now, from 1, we getar∆ABC = 4×12ar∆ABE = 2 ar∆ABE⇒ar∆ABE = 12ar∆ABC

In figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, prove that ar(ABE)=1/2ar(ABC).