1. **Problem statement:** Given two parallel lines $l \parallel m$, and a ray $\overrightarrow{DE}$ that bisects $\angle CDF$, find the value of $y$. 2. **Understanding the problem:** Since $l$ and $m$ are parallel, angles formed by a transversal have special relationships. 3. **Given:** $\angle CDE = 130^\circ$, $\overrightarrow{DE}$ bisects $\angle CDF$, and we want to find $y = \angle EDF$. 4. **Key fact:** The bisector divides $\angle CDF$ into two equal angles: $\angle CDE = \angle EDF = y$. 5. **Find $\angle CDF$:** Since $l \parallel m$, $\angle CDE$ and $\angle CDF$ are supplementary because they form a linear pair at point D. $$\angle CDE + \angle CDF = 180^\circ$$ Substitute $\angle CDE = 130^\circ$: $$130^\circ + \angle CDF = 180^\circ$$ $$\angle CDF = 180^\circ - 130^\circ = 50^\circ$$ 6. **Since $\overrightarrow{DE}$ bisects $\angle CDF$,** $$y = \frac{\angle CDF}{2} = \frac{50^\circ}{2} = 25^\circ$$ **Final answer:** $$y = 25^\circ$$