1. **Problem statement:** Given a figure with segments (AB) parallel to (CD) and (EF), angles BAC = 50° and CFE = ESC, find the measures of angles DCF and CFE. 2. **Step 1: Understand the parallel lines and transversal properties.** Since (AB) is parallel to (CD) and (EF), and FCA and ES are transversals, corresponding and alternate interior angles are equal. 3. **Step 2: Find angle DCF.** Angle BAC = 50° is given. Since (AB) is parallel to (CD), angle BAC corresponds to angle DCF (alternate interior angles). Therefore, $$\angle DCF = 50^\circ$$. 4. **Step 3: Find angle CFE.** Given that $$\angle CFE = \angle ESC$$. Since (EF) is parallel to (AB) and (CD), and ES intersects CD at S, angles CFE and ESC are equal by corresponding angles property. 5. **Step 4: Justify the measure of angle CFE.** Because $$\angle ESC = \angle CFE$$ and $$\angle DCF = 50^\circ$$, and considering the triangle formed by points C, F, E, the sum of angles in triangle CFE is 180°. 6. **Step 5: Show that (ES) is parallel to (FC).** Since $$\angle CFE = \angle ESC$$ and these are alternate interior angles formed by transversal FE crossing lines ES and FC, by the converse of the alternate interior angles theorem, (ES) is parallel to (FC). **Final answers:** $$\angle DCF = 50^\circ$$ $$\angle CFE = \angle ESC$$ (equal by given and parallel line properties) $$(ES) \parallel (FC)$$ by alternate interior angles theorem.