#### Answer

The path B,C,F,G,C,D,H,E,D,A,E,I,H,G,K,J,F,B is an Euler circuit.

#### Work Step by Step

Since all the vertices in this graph are even, the graph has zero odd vertices. Therefore, according to Euler's theorem, this graph has at least one Euler circuit.
To find an Euler circuit, we can start at any vertex.
Let's start at vertex B. According to Fleury's Algorithm, we should always choose an edge that is not a bridge, if possible. Since the edges BC and BF are not bridges, we can choose either of these edges as the next step in the path.
From vertex B, the path can travel to vertex C, then to vertex F, then to vertex G, then to vertex C, then to vertex D, then to vertex H, then to vertex E, and then to vertex D.
After this step, the path must travel to vertex A, then to vertex E, then to vertex I, then to vertex H, then to vertex G, then to vertex K, then to vertex J, then to vertex F, and then finally back to vertex B, because these are the only available edges.
This path is B,C,F,G,C,D,H,E,D,A,E,I,H,G,K,J,F,B. This path travels through every edge of the graph exactly once, so it is an Euler path. Since it starts and ends at the same vertex, this path is an Euler circuit.