16. In the figure below, $A B C$ is a triangle and $D$ is a point on side $B C$. Point $E$ is on side $A B$ such that $D E$ is the angle bisector of $\angle A D B$, and point $F$ is on side $A C$ such that $D F$ is the angle bisector of $\angle A D C$. Find the value of $\frac{A E}{E B} \cdot \frac{B D}{D C} \cdot \frac{C F}{F A}$.
16. Answer: 1
Since $D E$ is the angle bisector of $\angle A D B$, we have $\frac{A E}{E B}=\frac{A D}{B D}$. Similarly, since $D F$ is the angle bisector of $\angle A D C, \frac{A F}{C F}=\frac{A D}{D C}$. Hence $\frac{A E}{E B} \cdot \frac{B D}{D C} \cdot \frac{C F}{F A}=1$.