13. (15 points) As shown in the figure, in the isosceles right triangle $A B C$, $\angle A=90^{\circ}, A B=A C=1$, rectangle $E H G F$ is inside triangle $A B C$, and $G, H$ are on side $B C$. Find the maximum area of rectangle $E H G F$.
【Analysis】To maximize the area of rectangle $E H G F$, points $E$ and $F$ should be at the midpoints of $A B$ and $A C$, respectively. This is when the area of the rectangle is maximized.

【Solution】When the area of rectangle $E F G H$ is maximized, $E$ and $F$ should be on $A B$ and $A C$, respectively. Draw $B P \perp A B$ and $C P \perp A C$, and let $B P$ and $C P$ intersect at point $P$. Quadrilateral $A B P C$ is a square. Extend $E H$ to intersect $B P$ at $W$, and extend $F G$ to intersect $C P$ at $Q$. The sides $Q W$ and $A P$ intersect $E F$ at $U$ and $V$, respectively.
It is easy to prove that quadrilateral $E F Q W$ is a rectangle with vertices on the sides of square $A B P C$, and it is within square $A B P C$. Let $A E = x$, then $E B = 1 - x$,
$$
\begin{array}{l}
A E = A F = P W = P Q = x \\
E B = B W = F C = Q C = 1 - x
\end{array}
$$

Thus,
$$
E W = U V, \quad E U = A U, \quad W V = V P
$$

Therefore,
$$
E F + E W + W Q + Q F = A P + B C = \text{constant}
$$

That is, the perimeter of rectangle $E F Q W$ is constant. Among rectangles with the same perimeter, the one with the maximum area is a square. At this time, $A E : E B = 1$, so the maximum area of rectangle $E F G H$ is
$S_{\square E F G H} = \frac{1}{2} S_{\triangle A B C} = \frac{1}{4}$.
Answer: The maximum area of rectangle $E F G H$ is $\frac{1}{4}$.