1. **Problem statement:** Given that $\overline{AD} \parallel \overline{EG}$, $\overline{BH} \perp \overline{FC}$, and $m\angle ABH = 149^\circ$, find $m\angle EFH$. 2. **Identify known angles and relationships:** - Since $\overline{AD} \parallel \overline{EG}$, angles formed by a transversal with these lines have special relationships (alternate interior, corresponding angles). - $\overline{BH} \perp \overline{FC}$ means $\angle BHC = 90^\circ$. - $m\angle ABH = 149^\circ$ is given. 3. **Find $m\angle CBH$:** - At point B, angles $\angle ABH$ and $\angle CBH$ are adjacent and form a straight line (since points A, B, C are collinear). - Therefore, $m\angle CBH = 180^\circ - 149^\circ = 31^\circ$. 4. **Use right triangle $BHC$:** - Since $\overline{BH} \perp \overline{FC}$, $\angle BHC = 90^\circ$. - Triangle $BHC$ has angles $\angle CBH = 31^\circ$, $\angle BHC = 90^\circ$, so $\angle BCH = 180^\circ - 90^\circ - 31^\circ = 59^\circ$. 5. **Relate angles at point F:** - Since $\overline{AD} \parallel \overline{EG}$, and $\overline{FC}$ is a transversal, $\angle BCH = 59^\circ$ corresponds to $\angle EFH$ (alternate interior angles). 6. **Conclusion:** - Therefore, $m\angle EFH = 59^\circ$. **Final answer:** $$m\angle EFH = 59^\circ$$