Two opposite sides of a rectangle are each divided into $n$ congruent segments, and the endpoints of one segment are joined to the center to form triangle $A$. The other sides are each divided into $m$ congruent segments, and the endpoints of one of these segments are joined to the center to form triangle $B$. [See figure for $n=5, m=7$.] What is the ratio of the area of triangle $A$ to the area of triangle $B$?

$\text{(A)}\ 1\qquad\text{(B)}\ m/n\qquad\text{(C)}\ n/m\qquad\text{(D)}\ 2m/n\qquad\text{(E)}\ 2n/m$
Place the rectangle on a coordinate grid, with diagonal vertices $(0,0)$ and $(x,y)$.  Each horizontal segment of the rectangle will have length $\frac{x}{m}$, while each vertical segment of the rectangle will have length $\frac{y}{n}$.
The center of this rectangle will be $(\frac{x}{2}, \frac{y}{2})$.
Triangle $A$ has a base length of $\frac{y}{n}$, one of the vertical segments.  It has an altitude of $\frac{x}{2}$, which is the perpendicular distance from the center of the square to the left side of the square.  Thus, the area of triangle $A$ is $\frac{1}{2}\cdot\frac{y}{n}\cdot\frac{x}{2} = \frac{xy}{4n}$.
Triangle $B$ has a base length of $\frac{x}{m}$, one of the horizontal segments.  It has an altitude of $\frac{y}{2}$, which is the perpendicular distance from the center of the square to the bottom side of the square.  Thus, the area of triangle $B$ is $\frac{1}{2}\cdot\frac{x}{m}\cdot\frac{y}{2} = \frac{xy}{4m}$.
The ratio of areas is  $\frac{\frac{xy}{4n}}{\frac{xy}{4m}} = \frac{m}{n}$, which is answer $\boxed{B}$.