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Let a graph $G = (V, E) $ such that $|V| \geq 2$.
G is a connected graph, and deleting one vertex from G will not change the fact that it is connected (for any vertex). Let $u,v \in V(G)$ such that they are distinct vertices. Prove that G has a cycle that both $u$ and $v$ are part of.

I'm trying to prove this and need some help. Any graph gurus out there?

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    assume there is no cycle, that means there is only one path from $u$ to $v$. Remove an internal vertex and now they are disconected! – 2017-02-25
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    if $u$ and $v$ are adjacent, remove the edge, the graph must remain connected since its less than removing $u$ or $v$, so there is another path from $u$ to $v$ – 2017-02-25
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    I'm definitely not a graph Guru, but that seems simple enough: If there is no cycle containing $u$ and $v$, then all paths from $u$ and $v$ must have a common vertex. Now remove that vertex. What happens? – 2017-02-25
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    @cabo: There can definitely be more than one path from $u$ to $v$ (with the exception of the case of adjacent $u$ and $v$). They just have to have some vertex (other than $u$ and $v$) in common. – 2017-02-25
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    @celtschk; my bad, there can be many paths but a common vertex as you said. Thanks. – 2017-02-27

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