1. **Stating the problem:** We have trapezium ABCD with ABCE as a parallelogram and triangle ECD is isosceles with EC = ED. Given angles are $\angle A = 105^\circ$ and $\angle B = 39^\circ$. We need to find $\angle CDE$. 2. **Key properties:** - In parallelogram ABCE, opposite angles are equal and adjacent angles are supplementary. - Since ABCE is a parallelogram, $\angle A = \angle C$ and $\angle B = \angle E$. - Triangle ECD is isosceles with $EC = ED$, so $\angle ECD = \angle EDC$. 3. **Find $\angle C$ and $\angle E$:** - $\angle C = \angle A = 105^\circ$ - $\angle E = \angle B = 39^\circ$ 4. **Calculate $\angle CDE$:** - In triangle ECD, sum of angles is $180^\circ$: $$\angle ECD + \angle EDC + \angle CDE = 180^\circ$$ - Since $\angle ECD = \angle EDC = x$, and $\angle E = 39^\circ$ is adjacent to $\angle ECD$, we find: $$x + x + \angle CDE = 180^\circ \Rightarrow 2x + \angle CDE = 180^\circ$$ 5. **Relate $x$ to known angles:** - $\angle E = 39^\circ$ is part of parallelogram ABCE, so $\angle E + \angle CDE = 180^\circ$ (since $\angle CDE$ is adjacent to $\angle E$ on a straight line). - Therefore, $\angle CDE = 180^\circ - 39^\circ = 141^\circ$. 6. **Substitute $\angle CDE = 141^\circ$ into triangle ECD equation:** $$2x + 141^\circ = 180^\circ \Rightarrow 2x = 39^\circ \Rightarrow x = 19.5^\circ$$ 7. **Check consistency:** - $\angle CDE = 141^\circ$ is too large for the options given. 8. **Re-examine assumptions:** - Since ABCE is a parallelogram, $\angle E = 39^\circ$ and $\angle C = 105^\circ$. - Triangle ECD shares point C and D, with EC = ED. - The angle at E in triangle ECD is $\angle CED$, which is adjacent to $\angle B = 39^\circ$. 9. **Use the fact that $\angle CDE$ is the angle opposite side EC = ED:** - Since triangle ECD is isosceles with EC = ED, $\angle CDE = \angle ECD$. - Sum of angles in triangle ECD: $$2\angle CDE + \angle CED = 180^\circ$$ 10. **Find $\angle CED$:** - $\angle CED = 180^\circ - \angle B = 180^\circ - 39^\circ = 141^\circ$ (since ABCE is parallelogram and $\angle B + \angle E = 180^\circ$). 11. **Calculate $\angle CDE$:** $$2\angle CDE + 141^\circ = 180^\circ \Rightarrow 2\angle CDE = 39^\circ \Rightarrow \angle CDE = 19.5^\circ$$ 12. **Compare with options:** - Closest option is D: 27.5°. 13. **Conclusion:** The best approximate answer is $\boxed{27.5^\circ}$ (option D).