13. In $\triangle A B C$, it is known that $A B=9, B C: C A$ $=40: 41$. Find the maximum value of the area of $\triangle A B C$.
13. Let $AB = c$, $AC = br$, $BC = ar$, $(a < b)$. Establish a Cartesian coordinate system with point $A$ as the origin and the line $AB$ as the x-axis (as shown in the figure), set $A(0,0)$, $B(c, 0)$, $C(x, y)$.
From $\frac{BC}{AC} = \frac{a}{b}$, we get
$$
\frac{\sqrt{(x-c)^{2} + y^{2}}}{\sqrt{x^{2} + y^{2}}} = \frac{a}{b},
$$

Squaring both sides and simplifying, we get
$$
\left(x - \frac{b^{2} c}{b^{2} - a^{2}}\right)^{2} + y^{2} = \frac{a^{2} b^{2} c^{2}}{\left(b^{2} - a^{2}\right)^{2}}.
$$

This circle is called the Apollonian circle.
When the height of the triangle is the radius of the Apollonian circle, the area of the triangle is maximized. That is,
$$
S_{\triangle ABC} = \frac{1}{2} c \left(\frac{a b c}{b^{2} - a^{2}}\right).
$$

Substituting $\frac{a}{b} = \frac{40}{41}$ and $c = 9$ into the above formula, we get $S_{\triangle ABC} = 820$.