1. **Stating the problem:** We have triangle ABC with an exterior angle bisector AD' at vertex A. We are given that segment D'C is longer than AC and segment D'B is longer than AB. 2. **Understanding the exterior angle bisector theorem:** The exterior angle bisector theorem states that the bisector of an exterior angle of a triangle divides the opposite side externally in the ratio of the adjacent sides. Specifically, if AD' bisects the exterior angle at A, then: $$\frac{BD'}{D'C} = \frac{AB}{AC}$$ 3. **Given inequalities:** We have $D'C > AC$ and $D'B > AB$. This means point D' lies outside segment BC, extending it. 4. **Implications:** Since $D'C > AC$ and $D'B > AB$, the point D' lies on the line through B and C but outside segment BC, consistent with the exterior angle bisector theorem. 5. **Summary:** The exterior angle bisector AD' divides the opposite side BC externally in the ratio of the adjacent sides AB and AC, and the given inequalities confirm that D' lies outside segment BC. **Final answer:** The exterior angle bisector theorem applies, and the point D' lies outside segment BC such that: $$\frac{BD'}{D'C} = \frac{AB}{AC}$$