6. (3 points) Given a triangle $ABC$ with sides $AB=13, BC=20, AC=21$. On side $AB$, point $K$ is marked; on side $AC$, point $L$ is marked; on side $BC$, point $N$ is marked. It is known that $AK=4, CN=1, CL=\frac{20}{21}$. A line parallel to $NL$ is drawn through point $K$, intersecting side $AC$ at point $M$. Find the area of quadrilateral $NLMK$.
Answer: $\frac{493737}{11830}=41 \frac{8707}{11830}$

## Solution:

The area of triangle $ABC$ by Heron's formula is 126.

$$
\begin{gathered}
S_{CLN}=\frac{CN \cdot CL}{BC \cdot AC} \cdot S_{ABC}=\frac{1 \cdot \frac{20}{21}}{20 \cdot 21} \cdot 126=\frac{126}{21^{2}}=\frac{2}{7} \\
S_{BKN}=\frac{BK \cdot BN}{AB \cdot BC} \cdot S_{ABC}=\frac{19 \cdot 9}{13 \cdot 20} \cdot 126=\frac{10773}{130}
\end{gathered}
$$

Draw a line through point $B$ parallel to $NL$. It intersects side $AC$ at point $X$. Triangles $CLN$ and $CBX$ are similar, so $CX=\frac{CB \cdot CL}{CN}=\frac{20 \cdot \frac{21}{21}}{1}=\frac{400}{21}$, hence $AX=AC-CX=21-\frac{400}{21}=\frac{41}{21}$.

Triangle $MKA$ is similar to triangle $XBA$ with a ratio of $\frac{AL}{AB}=\frac{4}{13}$.

$$
\begin{gathered}
S_{MKA}=\frac{16}{169} S_{XBA}=\frac{16}{169} \cdot \frac{AX}{AC} \cdot S_{ABC}=\frac{16}{169} \cdot \frac{\frac{41}{21}}{21} \cdot 126=\frac{1312}{1183} \\
S_{NLMK}=S_{ABC}-S_{CLN}-S_{BKN}-S_{MKA}=126-\frac{2}{7}-\frac{10773}{130}-\frac{1312}{1183}=\frac{493737}{11830}=41 \frac{8707}{11830}
\end{gathered}
$$