1. **Stating the problem:** We have two parallel lines $\overline{HE}$ and $\overline{AD}$ intersected by a transversal $\overline{BF}$ at points $G$ and $C$. Given $m\angle HGF = 5n$ and $m\angle BCD = 2n + 66$, find $m\angle HGF$ and $m\angle FGE$. 2. **Key fact:** When a transversal crosses parallel lines, corresponding angles are equal. Here, $\angle HGF$ and $\angle BCD$ are corresponding angles, so: $$5n = 2n + 66$$ 3. **Solve for $n$:** $$5n - 2n = 66$$ $$\cancel{5n} - \cancel{2n} = 66$$ $$3n = 66$$ $$n = \frac{66}{3}$$ $$n = 22$$ 4. **Find $m\angle HGF$:** $$m\angle HGF = 5n = 5 \times 22 = 110$$ 5. **Find $m\angle FGE$:** Since $\overline{HE}$ is a straight line, angles $HGF$ and $FGE$ are supplementary: $$m\angle HGF + m\angle FGE = 180$$ Substitute $m\angle HGF = 110$: $$110 + m\angle FGE = 180$$ $$m\angle FGE = 180 - 110 = 70$$ **Final answers:** $$m\angle HGF = 110$$ $$m\angle FGE = 70$$