Question

In a equilateral triangle ABC the side BC is trisected at D prove that 7BC^2 = 9AD^2​

Answer 1

Soln:

Let the side of the equilateral triangle be "a"

and AE be the altitude of triangle ABC.

So, BE = EC =BC/2 = a/2.

and, AE = a√3/2.

Given that, BD = 1/3 BC = a/3.

So, DE = BD - BE

= a/2 - a/3

= a/6.

Now, applying Phytagorus Theorem in triangle ADE, we get,

AD^ = AE^ + DE^

= ( a√3/2)^ + (a/6)^

= (3a^/4) + a^/36

= 28a^/36

36AD^ = 28a^

9AD^ = 7AB^ ( "a" = AB,BC and AC)