$17 \cdot 47$ In triangle $ABC$, $\angle B$ is trisected by $BD$ and $BE$, intersecting $AC$ at $D$ and $E$, respectively, then
(A) $\frac{AD}{EC}=\frac{AE}{DC}$.
(B) $\frac{AD}{EC}=\frac{AB}{BC}$.
(C) $\frac{AD}{EC}=\frac{BD}{BE}$.
(D) $\frac{AD}{EC}=\frac{AB \cdot BD}{BE \cdot BC}$.
(E) $\frac{AD}{EC}=\frac{AE \cdot BD}{DC \cdot BE}$.
[Solution] Given that $B D$ bisects $\angle A B E$, then $\frac{A D}{D E}=\frac{A B}{B E}$, i.e., $A D=\frac{D E \cdot A B}{B E}$. Also, given that $B E$ bisects $\angle D B C$, then $\frac{D E}{E C}=\frac{B D}{B C}$, i.e., $E C=\frac{D E \cdot B C}{B D}$. Therefore, $\frac{A D}{E C}=\frac{D E \cdot A B}{B E} \cdot \frac{B D}{D E \cdot B C}=\frac{A B \cdot B D}{B E \cdot B C}$. Hence, the answer is $(D)$.