When $n$ is odd, $B$ has winning strategy. By induction - for $n=5$, it's obvious that $A$ loses in first move. If $n=2k+1$ for $k>2$, then $A$ in first move forms two polygons - one with $2l+1$ sides and other with $2m$, where $m>2$ and $m+l=k+1$. Now $B$ draws a diagonal such that $2m$-gon is divided into $2m-1$-gon (obviously not quadrilateral) and triangle. Now $2m-1$ and $2l+1$ are both smaller than $2k+1$, so $B$ can win in both of them by induction. So she just responds to a diagonal in the polygon it was drawn. When $n$ is even, $A$ has winning strategy. For $n=6$ she divides the hexagon to triangle and pentagon and $B$ loses in next move. For $n=2k$ with $k>3$, she divides the $n$-gon to two $k+1$-gons (she can because $k>3$) and plays symetrically. Q.E.D.