18. Fold a rectangular piece of paper in the manner shown in the figure, so that one vertex of the paper lands on the midpoint of one of the shorter sides. If the two shaded triangles are identical (all three sides are equal $\mathbf{a o}=\mathrm{do} ; \mathbf{o e}=\mathbf{o b} ; \mathbf{a e}=\mathbf{d b}$, all three angles are equal, and the areas are equal), and the length of the shorter side of the rectangular paper before folding is 12 cm. Then, the length of $\mathrm{BF}$ is? cm.
【Answer】10
Analysis: It is known that the two shaded triangles are identical, so $\mathrm{A}^{\prime} \mathrm{O}=\mathrm{DO}, \mathrm{EO}=\mathrm{B}^{\prime} \mathrm{O}, \mathrm{A}^{\prime} \mathrm{E}=\mathrm{B}^{\prime} \mathrm{D}$; since it is a fold, $\mathrm{BF}=\mathrm{B}^{\prime} \mathrm{F}, \mathrm{AB}=\mathrm{A}^{\prime} \mathrm{B}^{\prime}, \mathrm{AE}=\mathrm{A}^{\prime} \mathrm{E}$
Thus, it is easy to get $\mathrm{ED}=\mathrm{A}^{\prime} \mathrm{B}^{\prime}$, and $\mathrm{A}^{\prime} \mathrm{B}^{\prime}=\mathrm{AB}=12$ cm, i.e., $\mathrm{ED}=12$ cm
$\mathrm{DB}^{\prime}=\mathrm{B} ' \mathrm{C}=12 \div 2=6$ cm, $\mathrm{A}^{\prime} \mathrm{E}=\mathrm{B} ' \mathrm{D}=6$ cm, $\mathrm{AE}=\mathrm{A}^{\prime} \mathrm{E}=6$ cm,
$\mathrm{BC}=\mathrm{AD}=6+12=18$ cm, i.e., $\mathrm{B}^{\prime} \mathrm{F}+\mathrm{FC}=18$ cm
In $\mathrm{Rt} \triangle_{\mathrm{B}^{\prime} \mathrm{CF}}$, let $\mathrm{B}^{\prime} \mathrm{F}$ be $\mathrm{x}$ cm, then $\mathrm{FC}$ is $(18-\mathrm{x})$ cm, according to the Pythagorean theorem:
$x^{2}=6^{2}+(18-x)^{2}$, solving for $x$ gives $x=10$
$\mathrm{BF}=\mathrm{B}^{\prime} \mathrm{F}=10$ cm