Question

In an equilateral triangle ABC, AD is drawn perpendicular to BC meeting BC in A. Prove that AD^2=3BD^2.

Answer 1

$ad^{2} + bd^{2} = ab^{2} \\ ad^{2} + cd^{2} = ac^{2} \\ now \: ab = bc = ca \\ therefore \: \\ bd = cd = bc \div 2 \\ hence \: ad^{2} + (bc \div 2)^{2} = bc^{2} \\ 4 {ad}^{2} = 3 {bc}^{2}$