8. As shown in Figure 5, on the sides $AB$ and $BC$ of $\triangle ABC$, points $K$ and $M$ are taken such that $\frac{AK}{KB}=\frac{1}{4}, \frac{BM}{MC}=\frac{4}{5}$; on the line segment $KM$, point $O$ is taken such that $\frac{KO}{OM}=3$. $N$ is the intersection of ray $BO$ and $AC$, $AC=a$, and the distance from point $O$ to side $AC$ is $OD=d$. Then the area of $\triangle KMN$ is . $\qquad$
8. $\frac{a d}{3}$.

As shown in Figure 9, draw a line through point $M$ parallel to $B N$, intersecting $N C$ at point $M^{\prime}$; draw a line through point $K$ parallel to $B N$, intersecting $A N$ at point $K^{\prime}$, and connect $O M^{\prime}$ and $O K^{\prime}$.
Then $S_{\triangle N O M}=S_{\triangle O N M^{\prime}}$,
$S_{\triangle N O K}=S_{\triangle O N K}$.
Adding the two equations gives $S_{\triangle K U N}=S_{\triangle O W K}$.
Notice that, $\frac{N M^{\prime}}{M^{\prime} C}=\frac{B M}{M C}=\frac{4}{5}$.
Let $N M^{\prime}=4 u, M^{\prime} C=5 u$.
Since $\frac{K^{\prime} N}{N M^{\prime}}=\frac{K O}{O M}=3 \Rightarrow K^{\prime} N=12 u$.
Since $\frac{A K^{\prime}}{K^{\prime} N}=\frac{A K}{K B}=\frac{1}{4} \Rightarrow A K^{\prime}=3 u$.
Thus, $K^{\prime} M^{\prime}=12 u+4 u=16 u$,
$A C=3 u+12 u+4 u+5 u=24 u$.
Therefore, $\frac{K^{\prime} M^{\prime}}{A C}=\frac{16 u}{24 u}=\frac{2}{3}$
$\Rightarrow K^{\prime} M^{\prime}=\frac{2}{3} A C=\frac{2}{3} a$.
Hence, $S_{\triangle K U N}=S_{\triangle O M^{\prime} K^{\prime}}$
$=\frac{1}{2} K^{\prime} M^{\prime} \cdot O D=\frac{1}{2} \times \frac{2}{3} a d=\frac{a d}{3}$.