1. **State the problem:** We have two triangles, $\triangle ABC$ and $\triangle DEF$, with given angles and sides. We need to: a. Use the Triangle Angle Sum Theorem to find the unknown angles and justify the relationship between the triangles. b. Find the lengths of sides $AC$ and $DF$. 2. **Triangle Angle Sum Theorem:** The sum of the interior angles of any triangle is always $180^\circ$. That is, $$\text{angle}_1 + \text{angle}_2 + \text{angle}_3 = 180^\circ$$ 3. **Find the unknown angles:** - For $\triangle ABC$: Given $\angle A = 39^\circ$ and $\angle B = 25^\circ$, find $\angle C$: $$\angle C = 180^\circ - (39^\circ + 25^\circ) = 180^\circ - 64^\circ = 116^\circ$$ - For $\triangle DEF$: Given $\angle E = 116^\circ$ and $\angle D = 39^\circ$, find $\angle F$: $$\angle F = 180^\circ - (116^\circ + 39^\circ) = 180^\circ - 155^\circ = 25^\circ$$ 4. **Justify the relationship:** The angles of $\triangle ABC$ are $39^\circ, 25^\circ, 116^\circ$. The angles of $\triangle DEF$ are $39^\circ, 25^\circ, 116^\circ$. Since all corresponding angles are equal, the triangles are similar by the Angle-Angle (AA) similarity criterion. 5. **Find the lengths of $AC$ and $DF$:** Since the triangles are similar, corresponding sides are proportional: $$\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$$ Given: - $AB = 14$, $DE = 14$ - $BC = 20$ We need to find $AC$ and $DF$. Since $AB = DE = 14$, the scale factor is 1. Therefore, $AC = DF$. 6. **Find $AC$ using the Law of Cosines in $\triangle ABC$:** Law of Cosines formula: $$c^2 = a^2 + b^2 - 2ab \cos(C)$$ Here, $AC$ is opposite $\angle B = 25^\circ$, but to find $AC$, we can use sides $AB = 14$, $BC = 20$, and included angle $\angle B = 25^\circ$. So, $$AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(25^\circ)$$ Substitute values: $$AC^2 = 14^2 + 20^2 - 2 \times 14 \times 20 \times \cos(25^\circ)$$ $$AC^2 = 196 + 400 - 560 \times \cos(25^\circ)$$ Calculate $\cos(25^\circ) \approx 0.9063$: $$AC^2 = 596 - 560 \times 0.9063 = 596 - 507.53 = 88.47$$ $$AC = \sqrt{88.47} \approx 9.40$$ 7. **Find $DF$:** Since the scale factor is 1 ($AB = DE$), $$DF = AC \approx 9.40$$ **Final answers:** - $\angle C = 116^\circ$, $\angle F = 25^\circ$, triangles are similar. - $AC \approx 9.40$, $DF \approx 9.40$.