# Help with a proof

1. Dec 27, 2007

### Ynaught?

I am hoping someone can help me with a proof for the following conjecture:

If a triangle has sides with lengths (a,b,c) and sides (a,b) enclose an angle of 120 degrees, then:

a^2+ab+b^2=c^2

Mahalo

2. Dec 27, 2007

### dodo

Hey, Mahalo,
this one is easy.

In the attached picture, the 120-degree angle (on the green triangle) is marked with a black arc, while its 180-complement (on the red triangle), marked in yellow, would be a 60-degree angle.

On the red triangle, g^2 + h^2 = b^2 (eq.1).
On the bigger (red + green) triangle, (a + g)^2 + h^2 = c^2, or a^2 + 2ag + g^2 + h^2 = c^2 (eq.2).
Substituting (eq.1) into (eq.2), we get a^2 + 2ag + b^2 = c^2 (eq.3).
But, on the red triangle, g = b * cos(60) = b * 1/2; thus b = 2g, which turns (eq.3) into your equation.

#### Attached Files:

• ###### 120.png
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3. Dec 27, 2007

### HallsofIvy

Staff Emeritus
Dang! That's clever. I would have used the cosine law:
$c^2= a^2+ b^2- 2abcos(C)$
Since, here, C= 120 degrees, cos(C)= cos(120)= -1/2.

4. Dec 28, 2007

### Ynaught?

Thanks gents. I was trying to make things much too complicated. ;)