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31 lines
922 B
31 lines
922 B
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- 设$NF=x$,而且$AN+FD=NF$,所以$AD=BC=2x$
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- 对角线常见的辅助线引法，有$NH \perp BF$
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得到全等三角形$\triangle ANB \cong \triangle NHB$
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$\therefore BH=AB=3,NH=AN$
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- $\because \triangle BFE \cong \triangle BEC$
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$\therefore BF=BC=2x$
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又$\because BH=AB=3,\therefore HF=2x-3$
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<font color='red' size=4><b>
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相似三角形之反$A$形：$\triangle FNH \sim FNA$</b></font>
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$\therefore \frac{FN}{FB}=\frac{NH}{AB}=\frac{FH}{AF}$
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$\therefore \frac{x}{2x}=\frac{NH}{3}=\frac{2x-3}{AF}$
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$\therefore \frac{1}{2}=\frac{NH}{3}=\frac{2x-3}{AF}$
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$\therefore NH=\frac{3}{2}$
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$\therefore AN=\frac{3}{2}$
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$\therefore \frac{1}{2}=\frac{2x-3}{\frac{3}{2}+x}$
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$\therefore 4x-6=\frac{3}{2}+x$
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$3x=\frac{15}{2}$
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$\therefore x=\frac{5}{2}$
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$\therefore BC=2x=5$ |