Question

In an equilateral triangle ABC the side BC is trisected at D prove that 7BC2 = 9AD2​

Answer 1

Answer:

Let side =a

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

BE = a / 2 and BD = a /3

DE = BE - BD = a/ 6

In △ADE by Pythagoras theorem

AD^2 = AE^ 2 + DE ^2

Answer 2

Answer:

Check your answer please