2. Given that $D$ is a point on side $B C$ of $\triangle A B C$,
$$
\begin{array}{l}
\angle B=10^{\circ}, \angle C=20^{\circ}, \angle A D C=85^{\circ}, A B=10, \\
C D=m \text {. Then } B D=
\end{array}
$$
2. $\frac{10 m}{10-m}$.

As shown in Figure 5, on $AB$, take $AE = AC$, and connect $DE$.
From the given information, we have
$$
\begin{array}{l}
\angle BAD = \angle ADC - \angle B = 85^{\circ} - 10^{\circ} = 75^{\circ}, \\
\angle CAD = 180^{\circ} - \angle ADC - \angle C = 75^{\circ}.
\end{array}
$$

Thus, $\angle BAD = \angle CAD$.
Also, $AE = AC$, $AD = AD$, so $\triangle ADE \cong \triangle ADC$.
Therefore, $DE = CD = m$, $\angle AED = \angle C = 20^{\circ}$.
Hence, $\angle BDE = \angle AED - \angle B = 10^{\circ} = \angle B$.
So, $BE = DE = m$.
Thus, $AC = AE = AB - BE = 10 - m$.
By the Angle Bisector Theorem, $\frac{AB}{BD} = \frac{AC}{CD}$.
Therefore, $BD = \frac{AB \cdot CD}{AC} = \frac{10 m}{10 - m}$.