#### Answer

(a) This connected graph has no odd vertices. Therefore, by Euler's theorem, the graph has at least one Euler circuit.
(b) B,C,G,C,D,H,G,F,E,A,B,F,B is an Euler circuit.

#### Work Step by Step

(a) This connected graph has no odd vertices. Therefore, by Euler's theorem, the graph has at least one Euler circuit.
(b) Let's start at vertex B. From there, let's travel to vertex C, then to vertex G, then back to vertex C. From there, let's travel around the outside of the rectangle to vertex D, then to vertex H, then to vertex G, then to vertex F, then to vertex E, then to vertex A, and back to vertex B. There are only two edges which have not been used. The path can then travel to vertex F, and then finally back to vertex B.
This path is B,C,G,C,D,H,G,F,E,A,B,F,B. This path travels through every edge of the graph exactly once, so it is an Euler path. This Euler path starts and ends at the same vertex, so this path is an Euler circuit.
This is one Euler circuit but there are other Euler circuits in this graph also.