The Angle-Bisector 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 Angle-Bisector 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 Angle-Bisector 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 angle-bisector 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 6-8-10 triangle, so

*BZ*is 10.Next, set

*CU*equal to*x**.**UZ*then becomes 8 –*x*. Set up the angle-bisector 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).