How to Use the AngleBisector Theorem
The AngleBisector theorem states that if a ray bisects an angle of a triangle, then it divides the opposite side into segments that are proportional to the other two sides. The following figure illustrates this.
The AngleBisector theorem involves a proportion — like with similar triangles. But note that you never get similar triangles when you bisect an angle of a triangle (unless you bisect the vertex angle of an isosceles triangle, in which case the angle bisector divides the triangle into two congruent triangles).
Don’t forget the AngleBisector Theorem. (For some reason, students often do forget this theorem.) So whenever you see a triangle with one of its angles bisected, consider using the theorem.
How about an anglebisector problem? Why? Oh, just BCUZ.
Given: Diagram as shown
Find: 1.) BZ, CU, UZ, and BU and 2.) The area of triangle BCU and triangle BUZ

Find BZ, CU, UZ, and BU.
It’s a 6810 triangle, so BZ is 10.
Next, set CU equal to x. UZ then becomes 8 – x. Set up the anglebisector proportion and solve for x:
So CU is 3 and UZ is 5.
The Pythagorean Theorem then gives you BU:

Calculate the area of triangle BCU and triangle BUZ.
Both triangles have a height of 6 (when you use segment CU and segment UZ as their bases), so just use the triangle area formula:
Note that the ratio of the areas of these triangles, 9 : 15 (which reduces to 3 : 5), is equal to the ratio of the triangles’ bases, 3 : 5. This equality holds whenever a triangle is divided into two triangles with a segment from one of its vertices to the opposite side (whether or not this segment cuts the vertex angle exactly in half).