3. As shown in Figure 1, in Rt $\qquad$ $A B C$, $\angle B A C=90^{\circ}, A B=4$, $B C=12, A D \perp B C$ at point $D$, the angle bisector of $\angle A B C$ intersects $A C$ and $A D$ at points $E$ and $F$ respectively. A line through $F$ parallel to $B C$ intersects $A C$ at point $G$. Then the length of $F G$ is ( ).
(A) 10
(B) 6
(C) 8
(D) 9
3. C.

In the right triangle $\triangle ABC$, by the projection theorem, we have $AB^2 = BD \cdot BC$.
Therefore, $BD = \frac{AB^2}{BC} = \frac{4^2}{12} = \frac{4}{3}$. Hence, $DC = BC - BD = 12 - \frac{4}{3} = \frac{32}{3}$.
Since $BF$ is the angle bisector of $\angle ABC$, we have $\frac{AF}{FD} = \frac{AB}{BD} = 3$.
Thus, $\frac{AF}{AD} = \frac{AF}{AF + FD} = \frac{3}{4}$.
Since $FG \parallel CD$, we have $\frac{FG}{DC} = \frac{AF}{AD}$. Therefore, $FG = \frac{3}{4} \times \frac{32}{3} = 8$.