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- 设$NF=x$,而且$AN+FD=NF$,所以$AD=BC=2x$
- 对角线常见的辅助线引法，有$NH \perp BF$
得到全等三角形$\triangle ANB \cong \triangle NHB$
$\therefore BH=AB=3,NH=AN$

- $\because \triangle BFE \cong \triangle BEC$
  $\therefore BF=BC=2x$
  又$\because BH=AB=3,\therefore HF=2x-3$

<font color='red' size=4><b>
相似三角形之反$A$形：$\triangle FNH \sim FNA$</b></font>

$\therefore \frac{FN}{FB}=\frac{NH}{AB}=\frac{FH}{AF}$

$\therefore \frac{x}{2x}=\frac{NH}{3}=\frac{2x-3}{AF}$

$\therefore \frac{1}{2}=\frac{NH}{3}=\frac{2x-3}{AF}$

$\therefore NH=\frac{3}{2}$
$\therefore AN=\frac{3}{2}$

$\therefore \frac{1}{2}=\frac{2x-3}{\frac{3}{2}+x}$

$\therefore 4x-6=\frac{3}{2}+x$
$3x=\frac{15}{2}$

$\therefore x=\frac{5}{2}$

$\therefore BC=2x=5$