Show that a connected simple graph for which every edge is in some perfect matching is a block.
It seems obvious to prove by contradiction. so Suppose such a graph $G$ has a cut-vertex $v$.
The direction from here seems to be taking two components $H_1, H_2$ of $G-v$, and let $e_1, e_2$ be the edges by which $v$ was connected to those respectively.$e_1$ and $e_2$ are both in some perfect matchings in $G$, but they must be different matchings since they share a vertex. Call them $M_1, M_2$. How to continue from here? I'm guessing I'm looking for some contradiction to arise when looking at $M_1\cap E(H_2)$ or vice versa, but I'm not sure exactly how.