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Question
In an equilateral triangle ABC, the side BC is trisected at D. Prove that 9 AD2 = 7 AB2.
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Solution
Let side of equilateral triangle be a. And AE be the altitude of ΔABC
So, BE = EC = `"BC"/(2) = "a"/(2)`
And, AE = `("a"sqrt(3))/(2)`
Given that BD = `(1)/(3)"BC" = "a"/(3)`
So, DE = BD - BE = `"a"/(2) - "a"/(3) = "a"/(6)`
Now, in ΔADE by applying Pythagoras theorem
AD2 = AE2 + DE2
AD2 = `(("a"sqrt(3))/(2))^2 + ("a"/6)^2`
= `((3"a"^2)/4) + ("a"^2/36) = (28"a"^2)/(36)`
Or, 9 AD2 = 7 AB2.
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