Angle $B$ of triangle $ABC$ is trisected by $BD$ and $BE$ which meet $AC$ at $D$ and $E$ respectively. Then:
$\textbf{(A) \ }\frac{AD}{EC}=\frac{AE}{DC}  \qquad \textbf{(B) \ }\frac{AD}{EC}=\frac{AB}{BC} \qquad \textbf{(C) \ }\frac{AD}{EC}=\frac{BD}{BE} \qquad$
$\textbf{(D) \ }\frac{AD}{EC}=\frac{(AB)(BD)}{(BE)(BC)} \qquad \textbf{(E) \ }\frac{AD}{EC}=\frac{(AE)(BD)}{(DC)(BE)}$
Using the [Angle Bisector Theorem](https://artofproblemsolving.com/wiki/index.php/Angle_Bisector_Theorem) on $\angle ABE$ and $\angle DBC$, we find that $\frac{AD}{DE}=\frac{AB}{BE}$ and $\frac{DE}{EC}=\frac{BD}{BC}$. Rewriting this to better fit our answer choices gives $AD=\frac{(DE)(AB)}{BE}$ and $EC=\frac{(DE)(BC)}{BD}$. Hence, $\frac{AD}{EC}=\frac{(DE)(AB)}{BE}\cdot \frac{BD}{(DE)(BC)}\implies \boxed{\textbf{(D)}\ \frac{AD}{EC}=\frac{(AB)(BD)}{(BE)(BC)}}$.