Proof that the existence of a Hamilton Path in a bipartite graph is NP-complete
Given $G = (V,E)$, define $\tilde{G} = (\tilde{V},\tilde{E})$ by adding vertices $i^+$ and $i^-$ to $\tilde{V}$ for each $i\in V$ and edges $i^-j^+$ and $i^+j^-$ for each edge $ij\in E$.

It's not too hard to check that if $|V|$ is odd, $G$ has a Hamiltonian cycle if and only if $\tilde{G}$ is.  If $|V|$ is even, just add one new vertex of degree 2 with neighbors that have an edge between them.

So far, we've seen that Hamiltonian Cycle in bipartite graphs is hard.  To reduce this to Hamiltonian Path, just continue in the way you normally would to reduce Hamiltonian Cycle to Hamiltonian Path.