A connected k-regular bipartite graph is 2-connected. I've been struggling with this exercise; all ideas have been unfruitful, leading to dead ends. It is from Balakrishnan's A Textbook of Graph Theory, in the connectivity chapter:

Prove that a connected k-regular bipartite graph is 2-connected.

(That is, deletion of one vertex alone is not enough to disconnect the graph).
I think the objective is to make use of Whitney's theorem according to which a graph (with at least 3 vertices) is 2-connected iff any two of its vertices are connected by at least two internally disjoint paths. But I'll welcome any ideas or solutions.
Thank you!
 A: If there's a cut-vertex $v$, some neighbor of $v$ induces in $G \setminus \{v\}$ a connected bipartite graph with bipartition $A \cup B$.  It looks something like:

We observe:


*

*By definition, the number of edges from $A$ to $B$ is $k|V(A)| \equiv 0 \pmod k$.

*If $b$ is the number of edges from $B$ to $v$, then $b < k$, otherwise $v$ is not a cut vertex (in this case, deleting $v$ gives the connected graph induced by $A \cup B$).

*By definition, the number of edges from $B$ to $A$ is $k|V(B)|-b \not\equiv 0 \pmod k$, since $k \geq 2$.


This is a contradiction.
A: Let $G = (V_{1}\cup V_{2},E)$ be a connected, $k$-regular bipartite graph  where $V_{1}$ and $V_{2}$ are the partite vertex sets. As the case $k=1$ is trivial, we may assume that $k \geq 2$ and therefore $|V_{1}\cup V_{2}|\geq 4$.
Assume for contradiction that $G$ is not $2$-connected.
As $G$ is connected but not $2$-connected, there exists a vertex $v$ whose removal disconnects the graph. Without loss of generality we may assume that $v \in V_{1}$. Then $G-v = \uplus_{i\in [1,a]} G_{i}$ where each $G_{i}$ is a connected component and $a \geq 2$.
As $a \geq 2$, there exists some component $G_{b}$ such that $|V_{1}\cap V(G_{b})| \geq |V_{2}\cap V(G_{b})|$. (It shouldn't be too hard to convince yourself of this) For convenience denote $L = V_{1}\cap V(G_{b})$ and $R = V_{2}\cap V(G_{b})$.
As $G_{b}$ is a connected component and $G$ was connected, and $v \in V_{1}$, at least one vertex in $R$ was adjacent to $v$, and therefore has degree less than $k$. However the vertices in $L$ have lost no edges Then we have
$$
\sum_{u\in R}deg(u) < k\cdot|R| < k\cdot|L| = \sum_{w \in L}deg(w)
$$
However as $G[L \cup R]$ forms a bipartite graph, we know
$$
\sum_{u\in R}deg(u) = \sum_{w \in L}deg(w)
$$
Thus we have a contradiction, so $G$ must be at least $2$-connected.
A: Proof by contradiction.
Suppose $v$ is a cut vertex of $G = G(X,Y).$  Without loss of generality, let $v\in Y.$ Then $G-v$ has at least two components, say $C_1$ and $C_2.$ Let $(X_1,X_2)$ be the bi-partition of $C_1.$ Let $s$ be the number of neighbors of the vertex $v$ in $C_1$ (that is, in $X_1$), then $s<r.$
Since $C_1$ is bipartite, the number of edges incident at $Y_1$ in $C_1$ must be equal to the number of edges incident at $X_1$ in $C_1.$ Hence
$$ r|Y_1| = s(r-1) + (|X_1| - s)r$$
$$\Rightarrow  s = (|X_1| - |Y_1|)r$$
$$\Rightarrow  s \geq r,$$
a contradiction.
