Question

the line segment AB is produced to C so that AC = 3 AB an equilateral triangle BCD is drawn with BC as a side. prove that AD^2=7AB^2

Answer 1

Answer:

Let AB = x then AC = 3AB = 3x

and BC = 3x - x = 2x

Angle DBC = 60 ( equilateral triangle )

Angle DBA = 180 - 60 = 120°

So in ∆ABD use Cosine formula for angle B

Let AD = y

$\cos(120) = \frac{( {2x)}^{2} + {x}^{2} - {y}^{2} }{ {4x}^{2} } \\ \\ - \frac{1}{2} = \frac{4 {x}^{2} + {x}^{2} - {y}^{2} }{4 {x}^{2} } \\ \\ - {4x}^{2} = 8 {x}^{2} + 2 {x}^{2} - 2 {y}^{2} \\ 2{y}^{2} = 14 {x}^{2}\\y^2=7x^2$