In an equilateral ΔABC, D is a point on side BC such that BD=1/3BC. Prove that 9(AD)²=7(AB)²

Let E be the midpoint of BC so that BE=EC=BC/2 and the length of the sides of the equilateral triangle is x. So, we have AB=BC=CA=x…(1) BE=EC=x/2 And it is given that BD=BC/3=x/3 also BE=x/2 ∴DE=BE−BD=x/2−x/3=x/6 Now, In the right angle △AED we have AE=√3x/2 AD²=AE²+DE²=3x²/4+x²/36=28x²/36 36AD²=28x² Now put x²=AB² from equation (1), we get 36AD²=28AB² ⇒9AD²=7AB² [hence proved]