Let ABC be an equilateral triangle with point D on AC.
Create a new triangle with side lengths AD, DC, BD.
Prove one angle of this new triangle is 120 degrees.
I will let a=AD, b=BD, and c=CD. Then AB=BC=a+c. Also let T be the measure of angle BDC.
Apply the law of cosines to ADB and BDC:
(a+c)^2 = a^2+b^2-2ab*cos(180-T)
(a+c)^2 = b^2+c^2-2bc*cos(T)
Now I'll rearrange and simplify to isolate the cos() terms:
c(a+c)^2-a^2c-b^2c = 2abc*cos(T)
-a(a+c)^2+ab^2+ac^2 = 2abc*cos(T)
So now the right sides are identical. Then we can equate the left sides. After moving everything to one side:
c(a+c)^2-a^2c-b^2c +a(a+c)^2-ab^2-ac^2 = 0
(a+c)^3 - ac(a+c) - b^2(a+c) = 0
(a+c) * ((a+c)^2-ac-b^2)) = 0
a+c=0 can be rejected since we are working over positive quantities. Then (a+c)^2-ac-b^2 = 0 Rearrange to isolate b^2, and use the fact that cos(120)=-1/2:
b^2 = a^2 + c^2 + ac
b^2 = a^2 + c^2 - 2ac*cos(120)
The last equation is exactly the law of cosines for a triangle with sides of length a, b, and c with a 120 degree angle opposite of b.
Solution
