1. **Problem statement:** In triangle $\triangle ART$, line segment $SM$ is parallel to $AT$. Given some side lengths, find the indicated missing lengths using properties of similar triangles. 2. **Key concept:** Since $SM \parallel AT$, triangles $\triangle ASM$ and $\triangle ART$ are similar by the Basic Proportionality Theorem (Thales' theorem). This means corresponding sides are proportional: $$\frac{AS}{AR} = \frac{SM}{AT} = \frac{AM}{RT}$$ 3. **First problem:** Given $AS=3$, $RS=9$, $MT=4$, find $MR$. - Note that $AR = AS + RS = 3 + 9 = 12$. - Since $SM \parallel AT$, the segments on $RT$ are divided proportionally: $\frac{MR}{RT} = \frac{AS}{AR}$. - We know $MT=4$, so $RT = MR + MT$. Set up the proportion: $$\frac{MR}{MR + 4} = \frac{3}{12} = \frac{1}{4}$$ Cross-multiplied: $$4 \times MR = MR + 4$$ Simplify: $$4MR - MR = 4$$ $$3MR = 4$$ $$MR = \frac{4}{3}$$ 4. **Second problem:** Given $MR=8$, $MT=3$, $RS=7$, find $AS$. - $RT = MR + MT = 8 + 3 = 11$. - $AR = AS + RS$. - Using similarity: $$\frac{AS}{AS + 7} = \frac{MR}{RT} = \frac{8}{11}$$ Cross-multiplied: $$11 \times AS = 8 (AS + 7)$$ $$11AS = 8AS + 56$$ $$11AS - 8AS = 56$$ $$3AS = 56$$ $$AS = \frac{56}{3}$$ 5. **Third problem:** Given $AR=12$, $AS=5$, $RT=15$, find $MT$. - $RS = AR - AS = 12 - 5 = 7$. - Since $SM \parallel AT$, the segments on $RT$ are divided proportionally: $$\frac{MR}{RT} = \frac{AS}{AR} = \frac{5}{12}$$ - Let $MT = x$, then $MR = RT - MT = 15 - x$. Set up the proportion: $$\frac{15 - x}{15} = \frac{5}{12}$$ Cross-multiplied: $$12(15 - x) = 5 \times 15$$ $$180 - 12x = 75$$ $$180 - 75 = 12x$$ $$105 = 12x$$ $$x = \frac{105}{12} = \frac{35}{4} = 8.75$$ **Final answers:** - $MR = \frac{4}{3}$ - $AS = \frac{56}{3}$ - $MT = \frac{35}{4}$