Question

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

Answer 1

in equilateral triangle ABC, AD perpendicular to BC,

AD2 = AB2 - BD2 [ In triangle ABD, by Pythagoras Theorem]

= BC2 - BD2 [Since AB=BC]

= (2BD)2 - BD2

= 4BD2 - BD2

= 3BD2

Hence proved.