1. **Problem statement:** The perimeter of triangle $\triangle ABC$ is 29 m. The segment $\overline{AD}$ bisects angle $\angle A$. Given that $AB = 4$ m and $AC = 5$ m, find the lengths of $AB$ and $AC$. 2. **Understanding the problem:** Since $\overline{AD}$ bisects $\angle A$, by the Angle Bisector Theorem, the ratio of the sides opposite those angles is equal to the ratio of the adjacent sides. That is: $$\frac{BD}{DC} = \frac{AB}{AC}$$ 3. **Given:** - $AB = 4$ m - $AC = 5$ m - Perimeter $P = AB + BC + AC = 29$ m 4. **Find:** Lengths $AB$ and $AC$ (already given as 4 and 5 m respectively), but since the problem asks to find $AB$ and $AC$, we assume these are variables and the given 4 and 5 cm are labels on the diagram, so we clarify the problem as finding $AB$ and $AC$ given the perimeter and angle bisector. 5. **Let:** - $AB = x$ - $AC = y$ 6. **Using the Angle Bisector Theorem:** $$\frac{BD}{DC} = \frac{x}{y}$$ 7. **Since $D$ lies on $BC$, let:** - $BD = m$ - $DC = n$ Then $BC = m + n$. 8. **Perimeter equation:** $$x + y + (m + n) = 29$$ 9. **From the Angle Bisector Theorem:** $$\frac{m}{n} = \frac{x}{y} \implies m = \frac{x}{y} n$$ 10. **Substitute $m$ into perimeter:** $$x + y + \left( \frac{x}{y} n + n \right) = 29$$ $$x + y + n \left( \frac{x}{y} + 1 \right) = 29$$ 11. **Solve for $n$:** $$n = \frac{29 - x - y}{\frac{x}{y} + 1} = \frac{29 - x - y}{\frac{x + y}{y}} = \frac{29 - x - y}{\frac{x + y}{y}} = (29 - x - y) \cdot \frac{y}{x + y}$$ 12. **Then $m = \frac{x}{y} n = \frac{x}{y} \cdot (29 - x - y) \cdot \frac{y}{x + y} = (29 - x - y) \cdot \frac{x}{x + y}$ 13. **Since $BC = m + n$, sum is:** $$BC = (29 - x - y) \cdot \frac{x}{x + y} + (29 - x - y) \cdot \frac{y}{x + y} = (29 - x - y) \cdot \frac{x + y}{x + y} = 29 - x - y$$ 14. **This confirms the perimeter equation is consistent.** 15. **Given the problem states $AB = 4$ cm and $AC = 5$ cm, we can check if the perimeter matches:** $$4 + 5 + BC = 29 \implies BC = 20$$ 16. **Using the Angle Bisector Theorem:** $$\frac{BD}{DC} = \frac{4}{5}$$ 17. **Let $BD = 4k$ and $DC = 5k$, so:** $$BC = BD + DC = 4k + 5k = 9k$$ 18. **Since $BC = 20$, solve for $k$:** $$9k = 20 \implies k = \frac{20}{9}$$ 19. **Find $BD$ and $DC$:** $$BD = 4k = 4 \times \frac{20}{9} = \frac{80}{9} \approx 8.89$$ $$DC = 5k = 5 \times \frac{20}{9} = \frac{100}{9} \approx 11.11$$ 20. **Final answer:** - $AB = 4$ m - $AC = 5$ m These are consistent with the problem's given labels and the perimeter of 29 m.