4. In $\triangle A B C$, it is known that $A B=3, B C=6, A C=$ $4, D$ is a point on side $B C$, and $A D=3$. Then $\frac{B D}{D C}=(\quad)$.
(A) $\frac{7}{29}$
(B) $\frac{12}{29}$
(C) $\frac{29}{12}$
(D) $\frac{29}{7}$
4. D.

As shown in Figure 2, draw $A E \perp B C$ at point $E$. Let $B E=x$.

Then $C E=6-x$. Therefore, $3^{2}-x^{2}=4^{2}-(6-x)^{2} \Rightarrow x=\frac{29}{12}$ $\Rightarrow B D=\frac{29}{6}, C D=\frac{7}{6} \Rightarrow \frac{B D}{D C}=\frac{29}{7}$.