In an Equilateral Triangle Abc, D is a Point on Side Bc Such that Bd = `1/3bc` . Prove that 9 Ad^2 = 7 Ab^2 - Mathematics

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In an equilateral triangle ABC, D is a point on side BC such that BD = `1/3BC` . Prove that 9 AD2 = 7 AB2

 
 
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Solution

 

Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.

`∴ BE = EC = (BC)/2 = a/2`

And, AE = `(asqrt3)/2`

Given that, BD = `1/3BC`

∴ BD = a/3

`DE = BE - BD = a/2 - a/3 = a/6`

Applying Pythagoras theorem in ΔADE, we get

AD2 = AE2 + DE2

`AD^2 = ((asqrt3)/2)^2 + (a/6)^2`

`= ((3a^2)/4) + (a^2/36)`

`= (28a^2)/36`

`= 7/9 AB^2`

⇒ 9 AD2 = 7 AB2

 
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Chapter 6: Triangles - Exercise 6.5 [Page 151]

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