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In an equilateral triangle ABC, D is a point on side BC such that BD equals 1 3 BC. Prove that 9 AD square equals 7 AB square.

Q15   In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3  BC. Prove that
          9 AD^2 = 7 AB^2

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Given: An equilateral triangle ABC, D is a point on side BC such that BD = 1/3  BC.

To prove :  9 AD^2 = 7 AB^2

Let AB=BC=CA=a

Draw an altitude AE on BC.

So, BE=CE=\frac{a}{2}

In \triangleAEB, by Pythagoras theorem

AB^2=AE^2+BE^2

a^2=AE^2+(\frac{a}{2})^2

\Rightarrow a^2-(\frac{a^2}{4})=AE^2

\Rightarrow (\frac{3a^2}{4})=AE^2

\Rightarrow AE=(\frac{\sqrt{3}a}{2})

Given :  BD = 1/3  BC.

 BD=\frac{a}{3}

DE=BE=BD=\frac{a}{2}-\frac{a}{3}=\frac{a}{6}

In \triangleADE, by Pythagoras theorem,

AD^2=AE^2+DE^2

\Rightarrow AD^2=(\frac{\sqrt{3}a}{2})^2+(\frac{a}{6})^2

\Rightarrow AD^2=(\frac{3a^2}{4})+(\frac{a^2}{36})

\Rightarrow AD^2=(\frac{7a^2}{9})

\Rightarrow AD^2=(\frac{7AB^2}{9})

\Rightarrow 9AD^2=7AB^2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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