1. **State the problem:** We have two triangles, △ABC and △DEF, with given sides and angles: $AB=15$, $BC=20$, $\angle B=37^\circ$, $DE=24$, and $\angle E=37^\circ$. We want to find which additional side length (either $DF$ or $EF$) can be used to prove similarity by the Side-Angle-Side (SAS) similarity theorem. 2. **Recall the SAS similarity theorem:** Two triangles are similar if two sides of one triangle are proportional to two sides of the other triangle, and the included angles between those sides are equal. 3. **Identify the known angle and sides:** The given angles $\angle B$ and $\angle E$ are equal ($37^\circ$). The sides adjacent to these angles in each triangle are: - In △ABC: sides $AB=15$ and $BC=20$ adjacent to $\angle B$ - In △DEF: sides $DE=24$ and either $DF$ or $EF$ adjacent to $\angle E$ 4. **Check proportionality for each option:** - For $DF$ options (A and B), $DF$ is adjacent to $DE$ and $EF$ is adjacent to $DE$. - We want to find $DF$ or $EF$ such that the ratios of the sides around the equal angle are equal: $$\frac{AB}{DE} = \frac{BC}{DF}$$ or $$\frac{AB}{DE} = \frac{BC}{EF}$$ 5. **Calculate the ratio $\frac{AB}{DE}$:** $$\frac{15}{24} = \frac{5}{8} = 0.625$$ 6. **Test each option:** - Option A: $DF=8$ $$\frac{BC}{DF} = \frac{20}{8} = 2.5 \neq 0.625$$ - Option B: $DF=10$ $$\frac{20}{10} = 2 \neq 0.625$$ - Option C: $EF=29$ $$\frac{20}{29} \approx 0.6897 \neq 0.625$$ - Option D: $EF=32$ $$\frac{20}{32} = 0.625$$ 7. **Conclusion:** Only option D ($EF=32$) satisfies the SAS similarity condition because the ratios of the sides adjacent to the equal angle are equal. **Final answer:** The additional piece of information that could be used to prove similarity by SAS is $EF=32$.