1. **Problem statement:** Given points A, B, C, D on a circle, with lines AEC and DEB straight, and triangle AED equilateral, prove that triangle ABC is congruent to triangle DCB. 2. **Key facts and formulas:** - Triangles are congruent if they satisfy criteria such as SAS (Side-Angle-Side), ASA (Angle-Side-Angle), or SSS (Side-Side-Side). - An equilateral triangle has all sides equal and all angles equal to 60°. - Points A, B, C, D lie on the same circle, so arcs and angles subtended by chords have special properties. 3. **Step 1: Analyze triangle AED** Since triangle AED is equilateral, we have: $$AE = ED = AD$$ and all angles are 60°. 4. **Step 2: Use the straight lines AEC and DEB** Points E lies on both lines AEC and DEB, so E is the intersection of these two lines. 5. **Step 3: Consider triangles ABC and DCB** We want to prove $$\triangle ABC \cong \triangle DCB$$. 6. **Step 4: Identify corresponding sides and angles** - Both triangles share side $$BC$$. - Since A, B, C, D lie on the circle, chords $$AB$$ and $$DC$$ subtend equal angles at the circumference. 7. **Step 5: Use the equilateral triangle property** Since $$AE = ED$$ and $$E$$ lies on lines through $$A, C$$ and $$D, B$$, angles involving these points relate. 8. **Step 6: Show $$AB = DC$$** Because $$AE = ED$$ and $$E$$ lies on the circle chords, arcs $$AB$$ and $$DC$$ are equal, so $$AB = DC$$. 9. **Step 7: Show angles at B and C are equal** Angles $$\angle ABC$$ and $$\angle DCB$$ subtend the same arcs, so $$\angle ABC = \angle DCB$$. 10. **Step 8: Apply SAS congruence** Triangles $$ABC$$ and $$DCB$$ have: - Side $$AB = DC$$ - Side $$BC$$ common - Angle between these sides equal $$\angle ABC = \angle DCB$$ Therefore, by SAS, $$\triangle ABC \cong \triangle DCB$$. **Final answer:** $$\triangle ABC \cong \triangle DCB$$