16. As shown in Figure 5, in $\triangle A B C$, $A B=A C=\sqrt{5}$, $B C=2$, a circle $\odot O$ with $A B$ as its diameter intersects $A C$ and $B C$ at points $D$ and $E$ respectively. Then the area of $\triangle C D E$ is $\qquad$.
16. $\frac{2}{5}$.

Connect $A E$ and $B D$, and draw $D F \perp E C$ at $F$. Since $A B$ is the diameter of $\odot O$, we have $\angle A D B = \angle A E B = 90^{\circ}$. Also, since $A B = A C$, then $C E = \frac{1}{2} B C = 1$. Therefore, $A E = \sqrt{A C^{2} - C E^{2}} = 2$. From $\frac{1}{2} B C \cdot A E = \frac{1}{2} A C \cdot B D$, we know $B D = \frac{4 \sqrt{5}}{5}$. Thus, in $\triangle A B D$, $A D = \frac{3 \sqrt{5}}{5} \Rightarrow C D = \frac{2 \sqrt{5}}{5}$.
Since $\triangle C D F \sim \triangle C A E$, we have $\frac{C D}{C A} = \frac{D F}{A E}$. Solving this, we get $D F = \frac{4}{5}$.

Therefore, the area of $\triangle C D E$ is $\frac{1}{2} C E \cdot D F = \frac{2}{5}$.