1. **Problem Statement:** Explain why the given graph has at least one Euler path and find one such path starting at vertex A. 2. **Euler Path Criteria:** An Euler path is a path in a graph that visits every edge exactly once. 3. **Key Rule:** A graph has an Euler path if and only if it has exactly 0 or 2 vertices of odd degree (odd number of edges). 4. **Check Vertex Degrees:** - Vertex A connects to E, B, D twice (edges A-E, A-B, A-D, and D-A) so degree is 4 (even). - Vertex E connects to A, C, D (edges A-E, E-C, E-D) so degree is 3 (odd). - Vertex C connects to E, D (edges E-C, C-D) so degree is 2 (even). - Vertex D connects to C, B, A, E (edges C-D, D-B, A-D, E-D) so degree is 4 (even). - Vertex B connects to A, D (edges A-B, D-B) so degree is 2 (even). 5. **Count Odd Degree Vertices:** Only vertex E has an odd degree (3). 6. **Re-examine edges for A-D:** Edges listed include A-D and D-A, but these are the same edge, so count once. 7. **Correct Degrees:** - A: edges A-E, A-B, A-D = 3 (odd) - E: edges A-E, E-C, E-D = 3 (odd) - C: edges E-C, C-D = 2 (even) - D: edges C-D, D-B, A-D, E-D = 4 (even) - B: edges A-B, D-B = 2 (even) 8. **Odd Vertices:** Vertices A and E have odd degree (3 each). 9. **Conclusion:** Since exactly two vertices have odd degree, the graph has at least one Euler path. 10. **Euler Path Starting at A (Trial):** A, E, C, D, B, A, D, E 11. **Verify all edges used once:** Edges used: A-E, E-C, C-D, D-B, B-A, A-D, E-D 12. **Final Answer:** The graph has an Euler path because it has exactly two vertices of odd degree (A and E). One Euler path starting at A is: $$A, E, C, D, B, A, D, E$$