1. **Stating the problem:** We have two parallel lines $d_1$ and $d_2$ with points $A, B, C, D, E$ arranged such that angles $m(ABC) = 150^\circ$, $m(BCD) = 100^\circ$, and we want to find $m(CDE) = \alpha$. 2. **Relevant properties:** When two lines are parallel, alternate interior angles and corresponding angles have special relationships. Also, the sum of angles around a point or on a straight line is $180^\circ$. 3. **Analyzing the angles:** Since $d_1 \parallel d_2$, angles $m(ABC)$ and $m(CDE)$ are related through the transversal lines and the interior angles on the same side of the transversal sum to $180^\circ$. 4. **Calculate $\alpha$:** The sum of angles $m(BCD)$ and $m(CDE)$ is supplementary to $m(ABC)$ because they form a straight line or linear pair. So, $$m(ABC) + m(BCD) + m(CDE) = 360^\circ$$ But since $m(ABC) = 150^\circ$ and $m(BCD) = 100^\circ$, $$150 + 100 + \alpha = 360$$ $$\alpha = 360 - 250 = 110^\circ$$ 5. **Answer:** $\boxed{110^\circ}$ which corresponds to option D. ---