1. **State the problem:** In triangle $\triangle ART$, segment $SM$ is parallel to $AT$. Given some side lengths, find the indicated missing lengths using properties of similar triangles. 2. **Key concept:** When a segment is drawn parallel to one side of a triangle, it creates a smaller triangle similar to the original triangle. This means corresponding sides are proportional. 3. **Formula:** If $SM \parallel AT$, then by the Basic Proportionality Theorem (Thales' theorem), $$\frac{AS}{AR} = \frac{MR}{RT} = \frac{SM}{AT}$$ 4. **Given values for the first set:** $AS=3$, $MR=8$, $AR=12$. 5. **Find $RS$:** Since $AR=AS+RS$, then $$RS = AR - AS = 12 - 3 = 9$$ 6. **Check the ratio:** $$\frac{AS}{AR} = \frac{3}{12} = \frac{1}{4}$$ 7. **Given $MR=8$, find $RT$:** Using the ratio, $$\frac{MR}{RT} = \frac{1}{4} \Rightarrow RT = \frac{MR}{(1/4)} = 8 \times 4 = 32$$ 8. **For the second set:** Given $RS=9$, $MT=3$, $AS=5$. 9. **Find $AR$:** Since $AR=AS+RS$, $$AR = 5 + 9 = 14$$ 10. **Use the ratio:** $$\frac{AS}{AR} = \frac{5}{14}$$ 11. **Find $MR$ using $MT=3$ and $RT=MR+MT$:** $$\frac{MR}{MR + 3} = \frac{5}{14}$$ 12. **Solve for $MR$:** $$14 MR = 5 (MR + 3)$$ $$14 MR = 5 MR + 15$$ $$14 MR - 5 MR = 15$$ $$9 MR = 15$$ $$MR = \frac{15}{9} = \frac{5}{3}$$ 13. **For the third set:** Given $MT=4$, $RS=7$, $RT=15$. 14. **Find $AR$:** $$AR = AS + RS$$ 15. **Use the ratio:** $$\frac{AS}{AR} = \frac{MR}{RT}$$ 16. **Express $MR$ as $RT - MT = 15 - 4 = 11$** 17. **Set up the proportion:** $$\frac{AS}{AS + 7} = \frac{11}{15}$$ 18. **Solve for $AS$:** $$15 AS = 11 (AS + 7)$$ $$15 AS = 11 AS + 77$$ $$15 AS - 11 AS = 77$$ $$4 AS = 77$$ $$AS = \frac{77}{4} = 19.25$$ 19. **Summary of answers:** - $MR = \frac{5}{3}$ - $AS = 19.25$ - $MT = 4$ (given) **Final answers:** $$MR = \frac{5}{3}, \quad AS = 19.25, \quad MT = 4$$