A graph with Euler circuit - is it possible to get the circuit from every vertex in the graph?

Question

As mentioned in the topic, given a graph G, that contain Euler circuit. Does it mean that each vertex V in G can be the starting point of the Euler circuit? if so what is the proof? else is there any way to know from which vertex it starts?

Thanks.

Answer 1

Let $v_1v_2\cdots v_nv_1$ be one circuit (where some $v_i$ may be equal as the circuit goes through the same vertex multiple times). Then pick any $v_i$. Note note that $$ v_iv_{i+1}\cdots v_nv_1v_2\cdots v_{i-1}v_i $$ is also a circuit. So we can start the circuit wherever we like (in fact, not only any vertex, but any edge we want).