2. As shown in Figure 2, in quadrilateral $ABCD$, $AB=10$, $BC=17$, $CD=13$, $DA$ $=20$, $AC=21$. Then $BD=$
$2.10 \sqrt{5}$.
As shown in Figure 6, draw $B E \perp A C$ at point $E$, $D F \perp A C$ at point $F$, and draw $B G / / A C$ intersecting the extension of $D F$ at point $G$. Then, quadrilateral $B E F G$ is a rectangle, and $\triangle B D G$, $\triangle A B E$, and $\triangle C D F$ are all right triangles.
In $\triangle A B C$, we have
$$
p=\frac{1}{2}(A B+B C+C A)=24 \text{. }
$$

Thus, $B E=\frac{2}{21} \sqrt{24(24-10)(24-17)(24-21)}=8$.
Similarly, in $\triangle A D C$, we have $D F=12$.
Therefore, $D G=D F+F G=D F+B E=20$.
Also, $A E=\sqrt{A B^{2}-B E^{2}}=6$,
$$
C F=\sqrt{C D^{2}-D F^{2}}=5 \text{, }
$$

Then, $B G=E F=A C-A E-C F=10$.
Thus, $B D=\sqrt{D G^{2}+B G^{2}}=10 \sqrt{5}$.