Without sines. (Garrick this is for you)

Consider a triangle, ABC, and AE is the angle bisector of angle A. Now we extend the line AE, so that it reaches a point D, and that CD is parallel to AB. Since it is parallel to the line AD, we know that angle ABE = angle DCE, and angle EAB = angle EDC. Because 2 angles are the same, we know that by Angle angle similarity, they are similar triangles. Because they are similar triangles, we know that BA/BE = CD/CE Following that, we know that BAE is equal to CAE because they are split by an angle bisector. Now since angle CDE and CAE are equal, we know that it is an isosceles triangle, and that CD is equal to AC, substitute it back into the equation, getting us BA/BE = AC/CE, which is the angle bisector theorem.

With sines. (Everyone else.)

Again, we draw a triangle ABC, and we have AE is the angle bisector of angle A. We apply law of sines and get AB/BE = sin∠BEA/sin∠BAE, and that AC/EC = sin∠AEC/sin∠EAC. Since angle BEA and angle AEC are supplementary (Draw it you'll see), we know that the sin of angle BEA and the sin of angle AEC are equal. (Supplementary angles have equal sin's) Also since angle BAE and angle EAC are the same, (they were bisected), we know that they are also equal. Since they are the same, we can set the two ratios equal to each other, getting us AB/BE = AC/EC. After some algebra, shifting the variables around, this becomes the angle bisector theorem.