Equilateral triangle and 120° angle P is the angle's vertex. With A,B we form an equilateral triangle on the same direction the angle's rays open up.

How can I  prove the triangle's third vertex C is always on the bisector's line?
As pointed out by some comments, the problem has been originally misstated. It should be: we are given a 120 degrees angle with vertex P and an equilateral triangle ABC inscribed in it, such that the vertices A and B lie on the sides of the angle and C is in its interior, as in the drawing. Show that the line PC bisects the angle APB (i.e., that the angles APC and CPB are both  60 degrees).   

Proof. Consider  the quadrangle ACBP. It  is cocyclic (its vertices lie on the same circle) since the sum of the angles at its  opposite vertices  P and C is 180 degrees. Thus the angles APC and CPB are equal, since they are inscribed angles in the same circle that are facing equal segments. QED

From the proof you can see that it is enough to assume that ABC is isosceles, i.e. AC=BC, and that the sum of the angles and P and C is 180 degrees.
Call $\;M\;$ to the intersection point of $\;PC\,,\,AB\;$ , and observe that

$$\Delta APM\sim\Delta BPM\;\;\text{by s.a.s., since applying the angle bisector theorem we get}$$

$$\frac{AM}{MB}=\frac{AP}{PB}\;,\;\;\text{on the triangle}\;\;\Delta APB$$

This means $\;PC\perp AB\;$ (why?) , so both $\;\Delta CBP\;,\;\Delta CAP\;$ are right triangles (and, in fact, golden ones $\;30-60-90\;$).

Now just remember that an angle's bisector is the locus of all points on the plane that are equidistintant from both angle's rays.