If graph has no triangles then it is complete bipartite graph. Can anyone help me prove this theorem?

Let $G$ be a simple graph with $2n$ vertices and $n^2$ edges. If $G$
has no triangles, then $G$ is the complete bipartite graph $K_{n,n}$.

What I know: Assuming the $G$ has no triangles it follows that for every 3 vertices there are only 2 edges connecting them. For instance, having two sets of vertices i.e a's and b's. If I have vertices $a_1,b_1,b_2$. Then the edges are $a_1b_1,a_1b_2$. If I add another vertex i.e $a_2$, the edges added would be $a_2b_1,a_2b_2$. I added the edges such that there would be no triangles formed.
I get the idea of the complete bipartite graph through sketching it but I am having difficulty how can I come up to the definition of complete bipartite graph in order to prove the theorem.
Any insight is appreciated.
 A: Since $G=(V,E)$ has $2n$ vertices and $n^2$ edges, the degree-sum is $2n^2$ and the average degree is $n$, so the maximum degree is $\Delta(G)=n+t\ge n$. Choose a vertex $v$ of degree $\deg v=\Delta(G)=n+t$ where $t\ge0$. So $|N(v)|=\deg v=n+t$, and $|V\setminus N(v)|=n-t$.
Since $G$ is triangle-free, each vertex $w\in N(v)$ has neighbors only in $V\setminus N(v)$. It follows (since $G$ is a simple graph) that $\deg w\le n-t$. So $G$ has $n+t$ vertices of degree $\le n-t$, and the remaining $n-t$ vertices have degree $\le n+t$, so the degree sum is $\le2(n-t)(n+t)=2(n^2-t^2)$, and $|E|\le n^2-t^2$. Since $G$ has $n^2$ edges, we must have $t=0$, that is, $G$ is $n$-regular.
Let $V_1=N(v)$ and $V_2=V\setminus N(v)$, so $|V_1|=|V_2|=n$. Since $V_1$ is independent, and each vertex has degree $n$, it follows that every vertex in $V_1$ is joined to every vertex in $V_2$, and of course vice versa. There can be no other edges because $G$ is triangle-free. Hence $G$ is a complete bipartite graph $K_{n,n}$.
