Proof: First, let a=\frac{\angle A}{3}, ~~b=\frac{\angle B}{3}, ~~c=\frac{\angle C}{3}, then a+b+c = 60^{o} . Next, let a' = a + 60^{o}, ~b' = b + 60^{o}, ~c' = c + 60^{o}, then a + b' + c' = b + c' + a' = c + a' + b' = 180^{o}.
Now instead of starting with the triangle ABC, we start with an equilateral triangle A'B'C'. Next, outside the triangle A'B'C', we draw three triangles A'B'C, B'C'A, and C'A'B such that
\angle A'CB' = c, ~~\angle A'B'C= a', ~~\angle B'A'C = b',
\angle B'AC' = a, ~~\angle B'C'A = b', ~~\angle C'B'A = c',
\angle C'BA' = b, ~~\angle C'A'B = c', ~~\angle A'C'B = a' .
The reason we can do this is because a + b ' + c ' = b + c ' + a ' = c + a ' + b ' = 180^{o} as we have shown above.
Our job now is to prove
Our job now is to prove
\angle BAC' = \angle CAB' = a, ~~ \angle ABC' = \angle CBA' = b, ~~ \angle BCA' = \angle ACB' = c .
To prove this, we draw A'P and A'Q such that \angle C'A'P = 60^{o}, ~~\angle PA'B = c, ~~\angle B'A'Q = 60^{o}, ~~\angle QA'C = b. Clearly, the two triangles A'C'P and A'B'Q are equal and so we have A'P = A'Q.
Consider the two triangles BPA' and A'QC. These two triangles have two pairs of equal angles so they are similar triangles. Thus, \frac{BP}{A'Q} = \frac{BA'}{A'C}. Replacing A'Q=PA' we get \frac{BP}{PA'} = \frac{BA'}{A'C}. Combining this fact with \angle BPA' = \angle BA'C = 180^{o} - (b+c) we deduce that the two triangles BPA' and BA'C are similar triangles.
It follows that \angle A'BC = \angle PBA' = b, ~~~\angle A'CB = \angle PA'B = c.
It follows that \angle A'BC = \angle PBA' = b, ~~~\angle A'CB = \angle PA'B = c.
Prove similarly, we can obtain \angle BAC' = a, ~~~\angle ABC' = b and \angle CAB' = a, ~~~\angle ACB' = c, and thus the theorem is proved.
