3.

On the sides $AB, BC, CD$, and $AD$ of the convex quadrilateral $ABCD$, points $M, N, K$, and $L$ are located respectively, such that $AM: MB=3: 2, CN: NB=2: 3, CK=KD$, and $AL: LD=1: 2$. Find the ratio of the area of the hexagon $MBN KDL$ to the area of the quadrilateral $ABCD$.
Let's draw the diagonal $B D$ of the quadrilateral $A B C D$. Then,

$$
\begin{aligned}
& S_{\triangle C M X}=\frac{C N}{C B} \cdot \frac{C K}{C D} S_{\triangle B C D}=\frac{2}{5} \cdot \frac{1}{2} S_{\triangle B C D}=\frac{1}{5} S_{\triangle B C D}, \\
& S_{\triangle A M Z I}=\frac{A M}{A B} \cdot \frac{A L}{A D} S_{\triangle A B D}=\frac{3}{5} \cdot \frac{1}{3} S_{\triangle A B D}=\frac{1}{5} S_{\triangle A B D} .
\end{aligned}
$$

Therefore,

$$
\begin{gathered}
S_{M A B W X D L}=S_{A B C D}-S_{\triangle C V X}-S_{\triangle A M I}=S_{A B C D}-\frac{1}{5} S_{\triangle B C D}-\frac{1}{5} S_{\triangle A B D}= \\
=S_{A B C D}-\frac{1}{5}\left(S_{\triangle B C D}+S_{\triangle A B D}\right)=S_{A B C D}-\frac{1}{5} S_{A B C D}=\frac{4}{5} S_{A B C D}
\end{gathered}
$$

Thus,

$$
\frac{S_{M B N X D L}}{S_{A B C D}}=\frac{4}{5}
$$

## Answer

$\frac{4}{5}$