Example 5. As shown in Figure 6, $ABCD$ is a $2 \times 2$ square, $E$ is the midpoint of $AB$, $F$ is the midpoint of $BC$, $AF$ and $DE$ intersect at $I$, and $BD$ and $AF$ intersect at $H$. Find the area of quadrilateral $BEIH$.
Connect $AC$ intersecting $BD$ at $O$,
It is easy to prove $S_{\triangle BCD}=4, S_{\triangle BOD}$ $=S_{\triangle BOC}=S_{\triangle BFB}=1$.

Line $AHF$ intersects Rt $\triangle BOC$, by Menelaus' theorem we get
$$
\frac{BF}{FC} \cdot \frac{CA}{AO} \cdot \frac{OH}{HB}=1 \text {. }
$$

We can get $BH=\frac{2}{3} BO$.
Since $BO=OD$, we have $\frac{BH}{HD}=2$.
Thus, $S_{\triangle BFB}=\frac{2}{3} S_{\triangle OBF}=\frac{2}{3} \times \frac{1}{2} S_{BOC}$
$$
=\frac{1}{3} \text {. }
$$

Line $HIA$ intersects $\triangle EBD$, by Menelaus' theorem we get
$$
\frac{EI}{ID} \cdot \frac{DH}{HB} \cdot \frac{BA}{AE}=1 \text {. }
$$

We can also get $S_{\triangle EIA}=\frac{1}{5} S_{\triangle EBD}=\frac{1}{5}$.
Therefore, $S_{\mathrm{BEIH}}=S_{\triangle \mathrm{BFB}}-S_{\triangle \mathrm{AEI}}-S_{\triangle \mathrm{BFB}}$
$$
=1-\frac{1}{5}-\frac{1}{3}=\frac{7}{15} .
$$