bipartite-graphsgraph theory
Suppose $G$ is an undirected graph which is non-bipartite and is connected.
Is there a way to show that given two vertices $a$ and $b$, there exists a path of even length between the two ?
I tried to use the fact that there exist odd length cycles in a non-bipartite graph but couldn't use it to prove above statement .
Your idea is correct and doesn't have that many cases, you can simplify it quite a bit. Let $C$ be an odd cycle in $G$ of minimum length, this assures that $C$ isn't hamiltonian and there is a vertex $v\in V(G)\setminus V(C)$. As you said, there are 3 disjoint paths $t_1$, $t_2$ and $t_3$ between $v$ and $C$. Call $x_i$ the extreme of $t_i$ in $C$. Observe that for $i\neq j$, there are two $x_ix_j$-paths contained in $C$, one of even length and one of odd length. Define the cycle $C_{i,j}$ as follows:
Hence, the four odd cycles are $C$, $C_{1,2}$, $C_{1,3}$ and $C_{2,3}$.
The definition of bipartite graph at Wikipedia is:
... a bipartite graph ... is a graph whose vertices can be divided into two disjoint and independent sets $U$ and $V$ such that every edge connects a vertex in $U$ to one in $V$.
A bipartite graph doesn't need to be connected. It's fine to have $U$ and $V$ of any size (possibly even empty). We can also have any number of edges (possibly even zero).
Here's an example of a bipartite graph (with $U$ and $V$ indicated by vertex colors):
The graph drawn in the question is complete bipartite graph $K_{1,2}$, and we can highlight sets $U$ and $V$ as shown below:
Best Answer
Assuming a path allows crossing the same edges and vertices:
We are given vertices $a$ and $b$. We know that the graph contains a cycle of odd length. Let $c$ be a vertex on that cycle. Further, since the graph is connected, there exist paths $a\rightarrow c$ and $c\rightarrow b$. If the total length of the path $a\rightarrow c \rightarrow b$ is even, we are done. If not then consider the path $a\rightarrow c \rightarrow (\text{odd cycle back to $c$}) \rightarrow b$.