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三角形四心的概念介绍
(1)重心——中线的交点:重心将中线长度分成 2:1;
(2)垂心——高线的交点:高线与对应边垂直;
(3)内心——角平分线的交点(内切圆的圆心):角平分线上的任意点到角两边的距离相等;
(4)外心——中垂线的交点(外接圆的圆心):外心到三角形各顶点的距离相等.
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三角形四心与向量的结合
1.重心
`\overrightarrow {OA} + \overrightarrow {OB} + \overrightarrow {OC} = \overrightarrow 0 \Leftrightarrow ``O`是`\Delta ABC`的重心.
证明: 必要性是显然的, 下面证明充分性.
证法1:设 `O(x,y),A({x_1},{y_1}),B({x_2},{y_2}),C({x_3},{y_3})`
`\overrightarrow {OA} + \overrightarrow {OB} + \overrightarrow {OC} = \overrightarrow 0 \Leftrightarrow `
`\left\{ \begin{array}{l}({x_1} - x) + ({x_2} - x) + ({x_3} - x) = 0\\({y_1} - y) + ({y_2} - y) + ({y_3} - y) = 0\end{array} \right.`
`\Leftrightarrow \left\{ \begin{array}{l}x = \frac{{{x_1} + {x_2} + {x_3}}}{3}\\y = \frac{{{y_1} + {y_2} + {y_3}}}{3}\end{array} \right.`
证法二:
如图所示,因为\(\overrightarrow {OA} + \overrightarrow {OB} + \overrightarrow {OC} = 0\),
所以 \(\overrightarrow {OA} = - (\overrightarrow {OB} + \overrightarrow {OC} )\).
以\(\overrightarrow {OB} \),\(\overrightarrow {OC} \)为邻边作平行四边形\(BOCD\),
则有\(\overrightarrow {OD} = \overrightarrow {OB} + \overrightarrow {OC} \),
所以\(\overrightarrow {OD} = - \overrightarrow {OA} \).
又因为在平行四边形\(BOCD\)中,\(BC\)交\(OD\)于点\(E\),
所以 `\overrightarrow {BE} = \overrightarrow {EC} `,`\overrightarrow {OE} = \overrightarrow {ED}`.
所以\(AE\)是 的边\(BC\)的中线.
故\(O\)是$\vartriangle ABC$的重心.
2.垂心
`\overrightarrow {OA} \cdot \overrightarrow {OB} = \overrightarrow {OB} \cdot \overrightarrow {OC} = \overrightarrow {OC} \cdot \overrightarrow {OA} \Leftrightarrow O` 为 `\Delta ABC` 的垂心.
证明:如图所示 `O` 是三角形 `ABC` 的垂心,`BE` 垂直`AC`,`AD` 垂直`BC`, `D、E` 是垂足.
`\overrightarrow {OA} \cdot \overrightarrow {OB} = \overrightarrow {OB} \cdot \overrightarrow {OC} \Leftrightarrow \overrightarrow {OB} (\overrightarrow {OA} - \overrightarrow {OC} ) = \overrightarrow {OB} \cdot \overrightarrow {CA} = 0`
` \Leftrightarrow \overrightarrow {OB} \bot \overrightarrow {AC} `
同理 `\overrightarrow {OA} \bot \overrightarrow {BC} ` , `\overrightarrow {OC} \bot \overrightarrow {AB} `
` \Leftrightarrow ` `O` 为 `\Delta ABC` 的垂心.
3.内心
设 `a,b,c` 是三角形的三条边长,`O` 是 `\Delta ` ABC的内心 `a\overrightarrow {OA} + b\overrightarrow {OB} + c\overrightarrow {OC} = \overrightarrow 0 \Leftrightarrow O` 为 `\Delta ABC` 的内心.
证明: (1)证明必要性
`\because \frac{{\overrightarrow {AB} }}{c},\frac{{\overrightarrow {AC} }}{b}`分别为`\overrightarrow {AB} ,\overrightarrow {AC} `方向上的单位向量,
`\therefore \frac{{\overrightarrow {AB} }}{c} + \frac{{\overrightarrow {AC} }}{b}`平分`\angle BAC`,
` \because O` 为 `\Delta ABC` 的内心,
`\therefore \overrightarrow {AO} = \lambda (``\frac{{\overrightarrow {AB} }}{c} + \frac{{\overrightarrow {AC} }}{b}`),令`\lambda = \frac{{bc}}{{a + b + c}}`
`\therefore \overrightarrow {AO} = \frac{{bc}}{{a + b + c}}`(`\frac{{\overrightarrow {AB} }}{c} + \frac{{\overrightarrow {AC} }}{b}`)
化简得`(a + b + c)\overrightarrow {OA} + b\overrightarrow {AB} + c\overrightarrow {AC} = \overrightarrow 0 `
`\therefore a\overrightarrow {OA} + b\overrightarrow {OB} + c\overrightarrow {OC} = \overrightarrow 0 `.
(2)证明充分性
`\because a\overrightarrow {OA} + b\overrightarrow {OB} + c\overrightarrow {OC} = \overrightarrow 0 `
`\therefore a\overrightarrow {OA} b(\overrightarrow {OA}+\overrightarrow {AB})+c(\overrightarrow {OA}+\overrightarrow {AC})`,
化简得`(a + b + c)\overrightarrow {OA} + b\overrightarrow {AB} + c\overrightarrow {AC} = \overrightarrow 0 `
`\therefore \overrightarrow {AO} = \frac{{bc}}{{a + b + c}}`(`\frac{{\overrightarrow {AB} }}{c} + \frac{{\overrightarrow {AC} }}{b}`)
`\because \frac{{\overrightarrow {AB} }}{c},\frac{{\overrightarrow {AC} }}{b}`分别为`\overrightarrow {AB} ,\overrightarrow {AC} `方向上的单位向量,
`\therefore \frac{{\overrightarrow {AB} }}{c} + \frac{{\overrightarrow {AC} }}{b}`平分`\angle BAC`,
`\therefore O` 为 `\Delta ABC` 的内心.
4.外心
`\left| {\overrightarrow {OA} } \right| = \left| {\overrightarrow {OB} } \right| = \left| {\overrightarrow {OC} } \right|`` \Leftrightarrow ``O`为`\Delta ABC`的外心.
这个比较显然, 证明略.