- Question:
- In the figure above, ABC is a triangle. BD bisects the angle ABC. Prove AB/BC = AD/DC.

- Proof:
- Extended BD, and draw a line passing through point C and parallel to AB, the two lines intersects at point E.
- Since BA // CE (Made), so angle 1 = angle 3 (Two parallel lines BA and CE are cut by BE, then their alternate interior angles 1 and 3 are equal.)
- Since angle 1 = angle 2 (Given), and angle 1 = angle 3, so angle 2 = angle 3.
- In triangle BCE, since angle 2 = angle 3, so BC = CE (Property of an isosceles triangle.)
- In triangle ADB and triangle CDE,
- since angle 1 = angle 3, and angle ADB = angle CDE (Vertical angles are equal.)
- so triangle ADB is similar to triangle CDE (AA Theorem)
- so AB/CE = AD/DC (In similar triangles, corresponding sides are in proportion.)
- Since CE = BC, so AB/BC = AD/DC

**Note:**
In similar triangles ADB and CDE, AD opposite angle 1 and DC opposite angle 3, since angle 1 = angle 3, so AB and CE are a pair of corresponding sides.
Look another pair of corresponding sides, AB opposite angle ADB and CE opposite angle CDE, since angle ADB = angle CDE, so AB and CE are a pair
of corresponding sides. In two similar triangles, corresponding sides are in proportion. That is, AB/CE = AD/DC.

**What does this example tell us?**
In a triangle, the two line segments cut by an angle bisector and another two sides are in proportion.

Use this example to help middle school students learning inference.