II. (25 points) As shown in the figure, on the sides $AB$ and $CD$ of a square $ABCD$ with side length 1, points $M$ and $N$ are taken respectively. $AN$ and $DM$ intersect at $E$, and $BN$ and $CM$ intersect at $F$. Try to find the maximum area of quadrilateral $EMFN$, and indicate the positions of $M$ and $N$ when this maximum value can be achieved.
II. Solution First, consider the right trapezoid XYZU, whose upper and lower bases are $m, n$, and the right leg is 1. It is obvious that $\triangle X Y O$ and $\triangle U Z O$ have equal areas. Let their area be $t$, and the areas of $\triangle X O U$ and $\triangle Y O Z$ be $p, q$, respectively. It is easy to know that $\frac{t}{t}=\frac{U Q}{O Y}=$ $\frac{m}{n}=\frac{X O}{O Z}=\frac{t}{q}$, which means $p q=t^{2}$. Therefore, the heights of $\triangle X O U$ and $\triangle Y O Z$ are $\frac{m}{m+n}$ and $\frac{n}{m+n}$, respectively. Thus, we get $t=$ $\frac{m n}{2(m+n)}$. Now let $A M \cdots x, C N \cdots y$, then $M B \cdots 1 \cdots$. The area $=\frac{x(1-y)}{2(1+x-y)}, \angle M F N$ is $\frac{1-x}{2(1-y-x)}$, the combined area of $L F N E$ is $=\frac{x(1-y)}{2(1+x-y)}+\frac{y(1-x)}{2(1+y-x)}$ $=\frac{x(1-x)+y(1-y)}{2\left[1-(x-y)^{2}\right]} \geqslant 0$. However, the maximum value of this area is $\frac{1}{4}$. In fact, it is only necessary to have $2[x(1-x)+y(1-y)] \leqslant 1-(x$ $-y)^{2}$, which is equivalent to $(x+y-1)^{2} \geqslant 0$. Therefore, the maximum value of the area of quadrilateral MFNE is $\frac{1}{4}$, which is achieved when $x+y=1$ (i.e., $M$ is at the midpoint of $AB$).