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Question

In an equilateral triangle ABC, a point D is taken on base BC such that BD:DC=2:1. Prove that 9AD2=7AB2.

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Solution

Let A be origin and B(b) and C(c)
As triangle is equilateral
|b|=|c|=|bc|
AD2AB2=2c+b32|b|2=19(2c+b)(2c+b)|b|2
=19(4|c|2+4bc+|b|2)|b|2
{bc=|b||c|cosθ=|b|2cos60=|b|22}
AD2AB2=19|b|2|b|2(4+2+1)=79.

1214410_1395847_ans_05e2b00a56934c5395735cdb1864eac9.jpg

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