1. **Problem Statement:** Given that $AF \parallel HK$ and the angles at $H$ are $40^\circ$, $30^\circ$, and $30^\circ$, find the angles $\angle ABC$, $\angle ABH$, $\angle IHJ$, $\angle JHK$, $\angle DBH$, $\angle BHD$, $\angle HDB$, $\angle BDE$, $\angle HDF$, $\angle EDF$, and $\angle DFG$. 2. **Key Concept:** When two lines are parallel, corresponding angles and alternate interior angles are equal. Also, the sum of angles around a point is $360^\circ$, and the sum of angles in a triangle is $180^\circ$. 3. **Given Angles at $H$:** The angles around $H$ are $40^\circ$, $30^\circ$, and $30^\circ$. These sum to $100^\circ$, so the remaining angle at $H$ on the straight line is $180^\circ - 100^\circ = 80^\circ$. 4. **Find $\angle JHK$:** Since $AF \parallel HK$, and $JHK$ is on line $HK$, $\angle JHK = 40^\circ$ (corresponding to the $40^\circ$ angle at $H$). 5. **Find $\angle IHJ$:** The angle adjacent to $40^\circ$ and $30^\circ$ at $H$ is $30^\circ$, so $\angle IHJ = 30^\circ$. 6. **Find $\angle ABH$:** Since $AB$ is vertical and $HK$ is horizontal, and $AF \parallel HK$, $\angle ABH = 90^\circ$. 7. **Find $\angle ABC$:** Points $A$, $B$, and $C$ are vertical, so $\angle ABC = 90^\circ$. 8. **Find $\angle DBH$, $\angle BHD$, and $\angle HDB$:** Triangle $DBH$ has angles summing to $180^\circ$. Given the parallel lines and angles at $H$, $\angle DBH = 30^\circ$, $\angle BHD = 80^\circ$, and $\angle HDB = 70^\circ$. 9. **Find $\angle BDE$, $\angle HDF$, $\angle EDF$, and $\angle DFG$:** Using the intersecting lines and parallelism: - $\angle BDE = 40^\circ$ - $\angle HDF = 30^\circ$ - $\angle EDF = 110^\circ$ - $\angle DFG = 40^\circ$ **Final answers:** $$\angle ABC = 90^\circ$$ $$\angle ABH = 90^\circ$$ $$\angle IHJ = 30^\circ$$ $$\angle JHK = 40^\circ$$ $$\angle DBH = 30^\circ$$ $$\angle BHD = 80^\circ$$ $$\angle HDB = 70^\circ$$ $$\angle BDE = 40^\circ$$ $$\angle HDF = 30^\circ$$ $$\angle EDF = 110^\circ$$ $$\angle DFG = 40^\circ$$