In an equilateral triangle ABC, D is a point on the side BC such that BD = \(1\over 3\) BC. Prove that \(9{AD}^2\) = \(7{AB}^2\).

Solution :

Let ABC be an equilateral triangle and let D be a point on BC such thattriangle

BD = \(1\over 3\) BC.

Draw AE \(\perp\) BC. Join AD.

In \(\triangle\) AEB and AEC, we have :

AB = AC            (ABC is equilateral)

\(\angle\) AEB = \(\angle\) AEC

and AE = AE

\(\therefore\)  By SAS criteria of congruence, we have :

\(\triangle\) AEB \(\cong\) \(\triangle\) AEC

So,  BE = EC

Now, we have :

BD = \(1\over 3\)BC, DC = \(2\over 3\)BC and BE = EC = \(1\over 2\)BC                ……….(1)

Since, \(\angle\) C = 60, therefore

\(\triangle\) ADC is an acute triangle.

\(\therefore\)  \({AD}^2\) = \({AC}^2\) + \({DC}^2\) – \(2DC \times EC\)

= \({AC}^2\) + \(({2\over 3}BC)^2\) – \(2\times {2\over 3}BC\times {1\over 2}BC\)

= \({AC}^2\) + \(({4\over 9}BC)^2\) – \(({2\over 3}BC)^2\)

= \({AB}^2\) + \(({4\over 9}AB)^2\) – \(({2\over 3}AB)^2\)             (AB = AB = AC)

= \((9 – 4 – 6){AB}^2\over 9\) = \({7\over 9}{AB}^2\)

So,  \(9{AD}^2\) = \(7{AB}^2\)

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