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角分线定理

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发表于 2021-6-3 08:35 | 显示全部楼层 |阅读模式
本帖最后由 王守恩 于 2021-6-3 08:42 编辑

1,已知AD是角平分线,证明 \(\frac{AB}{AC}\equiv\frac{DB}{DC}\)

\(\displaystyle\frac{AB}{AC}\equiv\frac{\sin(b)}{\frac{\sin(b)\sin(a+b)}{\sin(b-a)}}\equiv\frac{\sin(b-a)}{\sin(a+b)}\equiv\frac{\sin(a)}{\frac{\sin(a)\sin(a+b)}{\sin(b-a)}}\equiv\frac{DB}{DC}\)

2,已知AD是角不均分线(平分只是特例,不均分才是普遍现象),

证明 \(\ \frac{AB*DC*\sin(x)}{AC*DB*\sin(y)}\equiv 1\ \ \ \    ∠BAC=x+y\)

\(\displaystyle\frac{AB*DC*\sin(x)}{AC*DB*\sin(y) }\equiv\frac{\sin(b)*\frac{\sin(y)\sin(x+b)}{\sin(b-y)}*\sin(x)}{\frac{\sin(b)\sin(x+b)}{\sin(b-y)}*\sin(x)*\sin(y)}\equiv 1\)
 楼主| 发表于 2021-9-8 12:34 | 显示全部楼层
本帖最后由 王守恩 于 2021-9-8 12:37 编辑

\(补充角分线定理与梅氏定理的关系。\)
\(O是△ABC内的点,\ D在AO的延长线上,\ E在BO的延长线上;\)
\(∠A=a_{1}+a_{2},\ \ \ ∠B=b_{1}+b_{2}\)
\(由\frac{AE*OB*\sin(a_{1})}{AB*OE*\sin(a_{2})}=1\Rightarrow\frac{AE*OB}{OE}=\frac{AB*\sin(a_{2})}{\sin(a_{1})}\ \ \ \ \ (1)\)
\(由\frac{AB*DC*\sin(a_{2})}{AC*DB*\sin(a_{1})}=1\Rightarrow\frac{DC}{AC*DB}=\frac{\sin(a_{1})}{AB*\sin(a_{2})}\ \ \ \ \ (2)\)
\((1)左边*(2)左边=(1)右边*(2)右边\)
\(\frac{AE*OB}{OE}*\frac{DC}{AC*DB}=\frac{AB*\sin(a_{2})}{\sin(a_{1})}*\frac{\sin(a_{1})}{AB*\sin(a_{2})}\Rightarrow\frac{AE*OB*DC}{AC*OE*DB}=1\)
\(同理。\)
\(由\frac{BD*OA*\sin(b_{1})}{BA*OD*\sin(b_{2})}=1\Rightarrow\frac{BD*OA}{OD}=\frac{BA*\sin(b_{2})}{\sin(b_{1})}\ \ \ \ \ (3)\)
\(由\frac{BA*EC*\sin(b_{2})}{BC*EA*\sin(b_{1})}=1\Rightarrow\frac{EC}{BC*EA}=\frac{\sin(b_{1})}{BA*\sin(b_{2})}\ \ \ \ \ (4)\)
\((3)左边*(4)左边=(3)右边*(4)右边\)
\(\frac{BD*OA}{OD}*\frac{EC}{BC*EA}=\frac{BA*\sin(b_{2})}{\sin(b_{1})}*\frac{\sin(b_{1})}{BA*\sin(b_{2})}\Rightarrow\frac{BD*OA*EC}{BC*OD*EA}=1\)
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