3. Point $D$ is inside $\triangle A B C$, given $A B=a b$, $B C=b c$, $C A=c a$, $A D=a d$, $B D=b d$, $C D=c d$ $(a, b, c, d$ are all positive numbers). Then $\angle A B D+\angle A C D=$ $\qquad$
$3.60^{\circ}$
As shown in Figure 5, construct $\triangle A C K \sim \triangle A B D$, then $\triangle A D K \sim \triangle A B C$.
At this point, $\frac{C K}{b d}=\frac{c a}{a b}$, which means
$$
C K=c d=C D \text {; }
$$

Similarly, $\frac{D K}{b c}=\frac{a d}{a b}$, which means
$$
D K=c d=C D \text {. }
$$

Therefore, $\triangle C D K$ is an equilateral triangle. Thus,
$$
\begin{array}{l}
\angle A B D+\angle A C D \\
=\angle A C K+\angle A C D=\angle D C K=60^{\circ} .
\end{array}
$$