Question

In an equilateral triangle ABC, the side BC is trisected at D. Prove that 9 AD² = 7 AB².

Answer 1

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.