Q5 of 53 Page 156

In equilateral ΔABC, D such that BD : DC = 1 : 2. Prove that 3AD = √7 AB.

BD : DC = 1 : 2


Let BD = x, DC = 2x –


⇒ AB = BC = AC = 3x


Let M be the mid point of BC


BM


∴ DM


⇒ DM


Applying Pythagoras theorem :


AB2 = BM2 + AM2


⇒ AM2


⇒ AM2


Applying Pythagoras theorem :


AD2 = AM2 + DM2


⇒ AD2


⇒ AD2 = 7x2


⇒ AD = √7x


So


AB : AD = 3x : √7x


⇒ AB : AD = 3 : √7


So 3AD = AB


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