"An angle bisector of an exterior angle of ?ABC intersects the line AB at point D. Prove that AD:BD=AC:BC."

To prove that \( AD : BD = AC : BC \), we can use the Angle Bisector Theorem and its extension to an exterior angle bisector.

Step 1: Identify the exterior angle at vertex C of \( \triangle ABC \) and its angle bisector. Let's call the exterior angle \( \angle ACE \) and the bisector intersects AB at point D.

Step 2: Apply the Exterior Angle Bisector Theorem, which states that the proportion of the lengths of the two segments created on one side of the triangle by the bisector is equal to the proportion of the lengths of the other two sides of the triangle.

So if CD is the bisector of the exterior angle at C, according to the Exterior Angle Bisector Theorem, we have:

\( \frac{AD}{DB} = \frac{AC}{BC} \)

The proof is a direct application of the theorem, and it doesn't require further geometric constructions or algebraic manipulations.