922 B
922 B
-
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
相似三角形之反$A$形：$\triangle FNH \sim FNA$
\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