Question

ΔABC is an equilateral triangle. Side BC is trisected at D. Prove that : 9AD² = 7AB².

Answer 1

Given: ABC is an equilateral triangle.

Side BC is trisected at D.

So, D lies on BC with `BD = DC = (BC)/3`.

To Prove: 9AD² = 7AB²

Proof (Step-wise)]:

1. Let AB = s

So, BC = s as well, since ABC is equilateral.

2. Place the triangle on a coordinate plane for convenience:

Take B = (0, 0), C = (s, 0). 

Then `A = (s/2, (ssqrt(3))/2)`.

3. Since D trisects BC,

`BD = s/3`, 

So, `D = (s/3, 0)`.

4. Compute AD² using the distance formula:

AD² = (x_A – x_D)² + (y_A – y_D)²

= `(s/2 - s/3)^2 + ((ssqrt(3))/2 - 0)^2`

5. Simplify:

`s/2 - s/3`

= `s(1/2 - 1/3)`

= `s(1/6)`

So, `(s/6)^2` 

= `s^2/36 xx ((ssqrt(3))/2)^2` 

= `(3s^2)/4`

= `(27s^2)/36` 

Therefore, `AD^2 = s^2/36 + (27s^2)/36`

= `(28s^2)/36`

= `(7/9) s^2`

6. Multiply both sides by 9:

`9AD^2 = 9 xx (7/9) s^2` 

= 7s² 

= 7AB²

Hence, 9AD² = 7AB², as required.