18. As shown in the figure, $E, F, G, H$ are the midpoints of the sides $AB, BC, CD, DA$ of quadrilateral $ABCD$, respectively. The intersection of $BH$ and $DE$ is $M$, and the intersection of $BG$ and $DF$ is $N$. What is $\frac{S_{\mathrm{BMND}}}{S_{\mathrm{ABCD}}}$?
18. As shown in the figure, $E, F, G, H$ are the midpoints of the sides $AB, BC, CD, DA$ of quadrilateral $ABCD$, respectively. The intersection of $BH$ and $DE$ is $M$, and the intersection of $BG$ and $DF$ is $N$. What is $\frac{S_{\mathrm{BMRD}}}{\mathrm{S}_{\mathrm{ABCD}}}$?
【Analysis】This problem makes full use of these midpoints. Connecting two midpoints, we get the midline. It is easy to see that triangles $ABD$ and $AEH$, and triangles $BDM$ and $HEM$ are similar. This not only gives the ratio of their areas but also the ratio of the sides needed. Here, we use triangle $EHM$ as an intermediate quantity to find the area relationships between triangles $BEM$, $EMH$, $NHD$, and $BDM$. From this, we can solve the problem.
【Solution】Solution: Connect $BD$, $EH$.
Since $E$, $H$ are midpoints,
$\therefore BD \parallel EH, BD = 2EH, \triangle EHM \sim \triangle DBM$
$\therefore S_{\triangle EHM}: S_{\triangle DBM} = 1: 4, \quad BM = 2HM$,
Assume $S_{\triangle EHM}$ is 1, then $S_{\triangle DBM}$ is 4.
$\therefore S_{\triangle EHM}: S_{\triangle BEM} = 1: 2$
Thus, $S_{\triangle BEM} = 2$,
Similarly, we get: $S_{\triangle DHM} = 2, S_{\triangle AEH} = S_{\triangle BEH} = 3$,
$\therefore S_{\triangle ABD}: S_{\triangle BDM} = 12: 4 = 3: 1$,
Similarly: $S_{\triangle BCD}: S_{\triangle BND} = 3: 1$,
$\therefore$ The area ratio of quadrilateral $BMDN$ to quadrilateral $ABCD$ is $1: 3$.
Therefore, $\frac{S_{\mathrm{BMRD}}}{S_{\mathrm{ABCD}}} = \frac{1}{3}$.