Question

In an equilateral triangle ABC the side BC is trisected at D. Prove that 9 AD = 7 AB?
​

Answer 1

Let side =a

Draw AE⊥BC So, as triangle ABC is equilateral.

BE= a/2 and as BD= a/3

DE=BE−BD=a/6

In △ADE by Pythagoras theorem,

AD^2 = AE^2+DE^2

=( √3/2 a)^2+ a^2 = 7a^2/9 = 7AB^2/9

= 9AD^2 =7AB^2.