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In an equilateral triangle A B C the ...

In an equilateral triangle `A B C` the side `B C` is trisected at `D` . Prove that `9\ A D^2=7\ A B^2`

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Let AB = BC = AC = 6x
Also ` BD = 1/3 BC = 1/3 xx 6x = 2x`
and ` BE = EC = (BC)/2 = 3x`
( perpendicular bisects the base in an equilatyeral triangle).
DE= BE- BD = 3x-2x=x
Now, by pythagoras theorem,
`AB^(2) = AE^(2) + BE^(2) = AD^(2)- DE^(2)+BE^(2)`
`(6x^(2)) = AD^(2)-x^(2)+ (3x)^(2)`
` Rightarrow " " AD^(2) = 36x^(2)+x^(2) - 9 x^(2) = 28x^(2)`
` 9AD^(2) = 9xx 28x^(2)= 9xx 7xx 4x^(2)= 7 xx 36x^(2) = 7 (AB)^(2)`
` 9AD^(2) = 7AB^(2)`
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