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Question

In an equilateral triangle ABC, the side BC is trisected at point D. Prove that 9AD2=7AB2. (Hint : Draw AEBC).
599649_3a03426de83f478bb73ccffddce4a3a1.png

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Solution


Let side =a
Draw AEBC So, as triangle ABC is equilateral.
BE=a2 and as BD=a3
DE=BEBD=a6
In ADE by Pythagoras theorem
AD2=AE2+DE2
=(32a)2+a362=7a92=7AB92
9AD2=7AB2

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