Probably this question was posted earlier here and no answer was given , also I have different point of doubt in my answer(different from him), so I'm posting this question.
Solution begins like:
For the converse , let $v$ be a cut vertex in a self complementary graph $G$. The graph $G − v$ has a spanning biclique, meaning a complete bipartite subgraph that contains all its vertices.
Since $G$ is self complementary, also $G$ must have a vertex $u$ such that $G − u$ has a spanning biclique.
Since each vertex of $G − v$ is nonadjacent to all vertices in the other components of $G − v$, Upto here it is fine .. i am getting it but i am not able to visualise the furthur points.
Since each vertex of $G − v$ is nonadjacent to all vertices in the other
components of $G − v$ , a vertex other than $u$ must be in the same partite set of the spanning biclique of $G − u$ as the vertices not in the same component as $u$ in $G − v$. Hence only $v$ can be in the other partite set, and $v$ has degree at least $n − 2$. We conclude that $v$ has degree at most $1$ in $G$, so $G$ has a vertex of degree at most $1$. Since a graph and its complement cannot both be disconnected, $G$ has a vertex of degree $1$.
Please help me out!! Thank you !