Hello, Riny!

I have a proof, but I don't know if it satisfies your needs . . .

In equilateral triangle ABC, D is a point on BC such that BD = (1/3)BC.

Prove that: .9·AD² = 7·AB² Code:

A
*
*:*
* : *
* : *
* : *
* : *
6 * : * 6
* : *
* : *
* : *
* : *
* :E 3 *
*-------*---*-----------*
B - 2 - D - - - 4 - - - C

Draw line segment AD.

Let the side of the triangle be 6: .AB = BC = CA = 6

. . Then: .BD = 2, .DC = 4

Draw altitude AE to side BC.

. . Then: .BE = EC = 3 and DE = 1

Triangle AEB is a 30__-__60 right triangle.

- - Hence: -AE = 3√3

In right triangle AED: .AD² .= .DE² + AE² .= .1² + (3√3)² .= .28

And we have: .9·AD² .= .9·28 .= .252

. . . - - - and: .7·AB² .= .7·6² .= .252

Therefore: .9·AD² .= .7·AB²