If v is the vertex, draw a circle about v, of any radius. Let it intersect the two rays in the points s and t. Draw two more circles, same radius, about s and t. These intersect in v, but they also intersect in u, inside the angle. Now the ray from v through u bisects the angle.
The Greeks wondered if an arbitrary angle could be trisected, i.e. cut into three equal pieces. Mathematicians worked on this problem for 2,000 years without success. Now we know why; it can't be done.
Let's begin by showing the cube root of 2 is not constructible. If it were, it would belong to a field extension of dimension 2m. Yet x3-2 has no rational roots, and is irreducible. Hence the cube root of 2 belongs to a field extension of dimension 3. Since 3 does not divide 2m, the cube root of 2 is not constructible. Similar reasoning holds for the 5th root of 7, and the 6th root of 38, and so on.
Now return to angle trisection. If 60 degrees can be trisected, then the coordinate x = cos(20) is constructible. Let c be the cosine of 20 degrees and let s be the sin, and refer to the triple angle formula. Note the formula for cos(3θ).
cos(3θ) = cos3(θ) - 3cos(θ)sin2(θ)
We know that the cosine of 60 is ½. Replace sin2 with 1-c2 and get this.
½ = c3 - 3c(1-c2)
1 = 8c3 - 6c
8c3 - 6c - 1 = 0
Try the rational roots with numerator ±1 and denominator 1 2 4 or 8. None of these work, hence the cubic is irreducible. The point c generates an extension of dimension 3, and is not constructible.
Another vexing problem was called "squaring the circle". Given a circle, find a square of equal area. This requires the construction of sqrt(π), which implies π, which isn't even algebraic. (We're not going to prove this here.) Transcendental numbers cannot be constructed, hence it is not possible to square the circle.