13. As shown in Figure 5, let the sum of the two sides $AC$ and $BC$ of $\triangle ABC$ be $a$, $M$ is the midpoint of $AB$, $MC=MA=5$. Then the range of values for $a$ is $\qquad$.
$$
13.10A B=10 \text {. }
$$

From $x+y=a$ and $x^{2}+y^{2}=10^{2}$, we can get $xy = \frac{a^{2}-10^{2}}{2}$. Therefore, $x$ and $y$ are the two real roots of the quadratic equation $z^{2}-a z+\frac{a^{2}-10^{2}}{2} = 0$. Hence, $\Delta=a^{2}-4 \times \frac{a^{2}-10^{2}}{2} \geqslant 0$, which means
$$
a \leqslant 10 \sqrt{2} \text {. }
$$