Thread: Similar Trianles,pythagoras

In An Equilateral Triangle -'abc', 'd' Is A Point On 'bc' Such That Bd=1/3bc .prove That 9ad^2 =7 Ab^2

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