1. **State the problem:** We have an isosceles triangle LMN with LM = LN. Lines AMB and CLN are parallel. DLM is a straight line. Angle $\angle DLN = 118^\circ$. We need to find the size of angle $\angle AMN$. 2. **Analyze the given information:** - Since LM = LN, triangle LMN is isosceles with $\angle LMN = \angle LNM$. - $DLM$ is a straight line, so $\angle DLM + \angle MLC = 180^\circ$. - Lines AMB and CLN are parallel, so alternate interior angles are equal. 3. **Find $\angle MLN$:** Since $\angle DLN = 118^\circ$ and $DLM$ is a straight line, then $$\angle MLN = 180^\circ - 118^\circ = 62^\circ.$$ 4. **Use isosceles triangle property:** In triangle LMN, since LM = LN, angles opposite these sides are equal: $$\angle LMN = \angle LNM.$$ Sum of angles in triangle LMN is $180^\circ$: $$\angle MLN + \angle LMN + \angle LNM = 180^\circ.$$ Substitute $\angle MLN = 62^\circ$ and $\angle LMN = \angle LNM = x$: $$62^\circ + x + x = 180^\circ,$$ $$2x = 180^\circ - 62^\circ = 118^\circ,$$ $$x = \frac{118^\circ}{2} = 59^\circ.$$ So, $$\angle LMN = \angle LNM = 59^\circ.$$ 5. **Use parallel lines property:** Since AMB and CLN are parallel, $\angle AMN$ and $\angle LNM$ are alternate interior angles. Therefore, $$\angle AMN = \angle LNM = 59^\circ.$$ **Final answer:** $$\boxed{59^\circ}.$$