Question

$In \: an \: equilateral \: triangle \: ABC, \\ \: D\: is \: a\: point\: on\: side \: BC \: such\\ \: that \: BD = \frac{1}{3} BC. Prove \\ that {9AD}^{2} = {7AB}^{2}$

please answer in form of equations...​

Answer 1

Step-by-step explanation:

proof : AB = BC = AC = X

in triangle AEB AND AEC

BE = EC by cpct

BE = EC = x/2

so , DE = x/6

using pythogeros theorem,

AE^2 = 3x^2/4

in triangle AED,

9AD^2 = 7AB^2