Given a simple connected bipartite graph $G$ with degree of vertices equal to $k$, where $k\ge 2$. Prove that there is no cut vertex exist in $G$.

Question

Given a simple connected bipartite graph $G$ with degree of vertices equal to $k$, where $k\ge 2$. Prove that there is no cut vertex exist in $G$.

Cut vertex $v$ here is a vertex which make the graph induced have number of connected component $>1$ when $v$ is removed.

I have tried to prove by contradiction but i have no clue about what contradiction can be obtained. I am quite curious about what the point the graph has to be bipartite is here. I have not come across any place to adopt the property of a bipartite graph so far. Any hints on tackling this problem would be appreciated. Last but not least, thanks for reading.

Edited: i have found something

Answer 1

Hint:

Show that the number of edges in the original graph is a multiple of $k$.

Show that the removal of one vertex removes $k$ edges.

Show that the number of edges in each component (after the removal) still is a multiple of $k$.

Conclude that the original graph must already have been disconnected.

Answer 2

Let $x$ be a cut vertex and properly $2$-color the graph red and green. If we delete $x$ we get some connected components: in one of these components, let the red vertices belong to $A$ and green vertices belong to $B$.

The graph induced by $\{x\} \cup A \cup B$ looks as follows:

We reach a contradiction if we compare modulo $k$ (a) the number of edges coming out of $A$ and (b) the number of edges going into $A$.