ABCD

K

AB

L

BC

M

CD

N

AD

\frac{AK}{KB}=2

\frac{BL}{LC}=\frac{1}{3}

\frac{CM}{MD}=1

\frac{DN}{NA}=\frac{1}{5}

AKLCMN

\frac{11}{12}

S_{\triangle BKL}=\frac{BK}{AB}\cdot\frac{BL}{BC}\cdot S_{\triangle ABC}

AC

S_{\triangle BKL}=\frac{BK}{AB}\cdot\frac{BL}{BC}\cdot S_{\triangle ABC}=\frac{1}{12}\cdot S_{\triangle ABC},

S_{\triangle DMN}=\frac{DN}{AD}\cdot\frac{DM}{DC}\cdot S_{\triangle ADC}=\frac{1}{12}\cdot S_{\triangle ADC},

S_{\triangle BKL}+S_{\triangle DMN}=\frac{1}{12}\cdot(S_{\triangle ABC}+S_{\triangle ADC})=\frac{1}{12}\cdot S_{ABCD}=\frac{1}{12},

S_{AKLCMN}=S_{ABCD}-(S_{\triangle BKL}+S_{\triangle DMN})=1-\frac{1}{12}=\frac{11}{12}.