AHSME 1952 Problem 19

Angle $B$ of triangle $ABC$ is trisected by $BD$ and $BE$ which meet $AC$ at $D$ and $E$ respectively. Then: $\textbf{(A) \ }\frac{AD}{EC}=\frac{AE}{DC} \textbf{(B) \ }\frac{AD}{EC}=\frac{AB}{BC} \textbf{(C) \ }\frac{AD}{EC}=\frac{BD}{BE}$ $\textbf{(D) \ }\frac{AD}{EC}=\frac{(AB)(BD)}{(BE)(BC)} \textbf{(E) \ }\frac{AD}{EC}=\frac{(AE)(BD)}{(DC)(BE)}$