11. As shown in the figure, square $\mathrm{ABCD}$ and square $\mathrm{EFGH}$, their four pairs of sides are parallel to each other. Connect $\mathbf{C G}$ and extend it to intersect $\mathbf{B D}$ at point $\mathbf{I}$. Given that $B D=10$, the area of triangle $B F C$ is $3$, and the area of triangle $\mathrm{CHD}$ is $5$, then the length of $B I$ is?
【Answer】 $\frac{15}{4}$
Analysis:
Method one: Equal area transformation + Swallowtail model
Connect $\mathrm{BG}, \mathrm{DG}, \mathrm{FG} \| \mathrm{BC}, \mathrm{S}_{\triangle \mathrm{BCG}}=\mathrm{S}_{\triangle \mathrm{BCF}}=3$,
Similarly, $\mathrm{S}_{\triangle \mathrm{CDG}}=\mathrm{S}_{\triangle \mathrm{CDH}}=5$,
$\mathrm{BI}: \mathrm{DI}=\mathrm{S}_{\triangle \mathrm{BCG}:} \mathrm{S}_{\triangle \mathrm{CDG}}=3: 5$,
$\mathrm{BI}=10 \div(3+5) \times 3=\frac{15}{4}$
Method two: $\square$
Draw perpendiculars through I, G, F, H, then $\mathrm{FJ}=\mathrm{GN}, \mathrm{HK}=\mathrm{GO}$
Let the side length be $\mathrm{a}$, then $2 a^{2}=100$. By the triangle area formula, we get $H K=\frac{a}{5}, \quad F J=\frac{3}{25} a$
$\mathrm{IL}$ and IM are proportional enlargements of $\mathrm{GN}$ and $\mathrm{GO}$, respectively, let both be enlarged by a factor of $\mathrm{k}$
$$
\frac{1}{2} a^{2}=s_{A B C}=s_{B I C}+s_{C D I}=\frac{1}{2}(B C \times I L+C D \times I M)
$$

So $k=\frac{25}{8}, \mathrm{IL}$ equals $\mathrm{FJ}$ enlarged by $\mathrm{k}$ times, which is $\frac{5 a}{8}$,
$$
\begin{array}{l}
B I^{2}=2 I L^{2}=2 \times \frac{5^{2}}{8^{2}} \times \frac{10^{2}}{2}=\frac{5^{2} \times 10^{2}}{8^{2}} \\
B I=\frac{25}{4}
\end{array}
$$