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28 lines
1.2 KiB
28 lines
1.2 KiB
2 years ago
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- 给出离的很远的两条边相等，就是让我们证明两个三角形全等：
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$\because BG=EH,\angle ABG=\angle BEH,AB=BE$
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$\therefore \triangle ABG \cong \triangle BHE$
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目的：
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$\angle 1=\angle 2,\angle 3=\angle 4$
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$\because \angle 1+\angle 3=90^{\circ},\angle 1=\angle 2 \Rightarrow \angle 2+\angle 3=90^{\circ}$
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$\therefore \angle BOA=90^{\circ}$
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- 这里有一个 **子母型** 的相似直角三角形:
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$\triangle AOB \sim \triangle ABG$
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$\therefore \frac{OB}{OA}=\frac{BG}{AB}=\frac{2}{5}$
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因为知道$\triangle NOM$是直角三角形，所以应该是需要继续找出相似的直角三角形。
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利用两个相交的直角，可以有$\angle AOM+\angle 5=\angle AOM+\angle 6$
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$\therefore \angle 5=\angle 6$
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$\because \angle 7=\angle 3,\angle 8=\angle 4$
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$\because \angle 3=\angle 4$
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$\therefore \angle 7=\angle 8$
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$\therefore \triangle OMB \sim \triangle OAN$
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$\therefore \frac{BM}{AN}=\frac{OB}{OA}=\frac{2}{5}$
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$\therefore BM=\frac{2}{5}*\frac{5}{2}=1$
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$\therefore AM=5-1=4$
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$\therefore tan \angle AMN=\frac{AN}{AM}=\frac{5}{2} ÷ 4 =\frac{5}{8}$
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