Friendship theorem: need help with part of proof. Suppose $G$ is a simple graph such that every two of its vertices have exactly one common neighbor.
The friendship theorem says that $G$ must be a friendship graph (a bunch of triangles joined at a single vertex)

The hint in the problem says to suppose for a contradiction that $G$ is not such a graph. Then supposedly $G$ must be regular (all vertices have same degree), but I don't see why.  Can someone help? Once I have this, I see how to get the contradiction to finish the proof.
EDIT: I posted my solution. If you are interested, please let me know if it looks ok. If you have a simpler proof feel free to post it, I'd like to see it.
 A: Suppose every two vertices in $G$ have exactly one common neighbor, but $G$ is not a friendship graph. We show $G$ is regular.
Let $u$ and $v$ be vertices of $G$. We define an injection $f:N(u)\to N(v)$, yielding $|N(u)|\leq |N(v)|$. By symmetry the reverse inequality will follow.
Case 1: $v\notin N(u)$. Let $f(x)$ be the neighbor between $x$ and $v$. If $x\neq y$ then by uniqueness of common neighbors $f(x)=f(y)$ would imply $u=f(x)$, but then $v\in N(u)$.
Case 2: $v\in N(u)$. Since $G$ is not a friendship graph, i.e. it has no vertex that is connected to all other vertices, there is a vertex $w\notin N(u)$. For each $x\in N(u)$ let $x'$ be the vertex between $x$ and $w$. Define $f(v)=v'$ and for all other $x$ let $f(x)$ be the vertex between $x'$ and $v$. 
If $x\neq v\in N(u)$ but $f(x)=f(v)$, then by uniquess of the common neighbor of $x$ and $v$ we have $u=f(v)\in N(w)$, a contradiction.
If $x\neq y\in N(u)$ with $x\neq v\neq y$ but $f(x)=f(y)$, then by uniqueness of the common neighbor of $x'$ and $y'$ we have $w=v\in N(u)$, a contradiction.
