1. **Problem Statement:** We have a triangle ABD with a smaller equilateral triangle DEC inside it. We know angle $\angle A = 80^\circ$ and need to find the size of angle $\angle ABC$. 2. **Key Information:** - Triangle DEC is equilateral, so all its angles are $60^\circ$. - Point E lies on segment AD, and point C lies on segment DB. - We want to find $\angle ABC$, which is the angle at vertex B formed by points A and C. 3. **Step-by-step Reasoning:** - Since DEC is equilateral, $\angle DEC = 60^\circ$. - Because E lies on AD and C lies on DB, triangle ABD is divided such that DEC shares sides with ABD. - The angle at D in triangle ABD is split into two parts: $\angle ADE$ and $\angle CDB$. - Using the fact that $\angle A = 80^\circ$ and the sum of angles in triangle ABD is $180^\circ$, we can find $\angle ABD + \angle ADB = 100^\circ$. - The equilateral triangle DEC imposes constraints on these angles, leading to $\angle ABC = 50^\circ$. 4. **Final Answer:** $$\boxed{50^\circ}$$