Cho tam giac ABC. Tim M thoa : | vecto MA + vecto BC| = |vecto MC + vecto BC|

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Cho tam giac ABC. Tim M thoa : | vecto MA + vecto BC| = |vecto MC + vecto BC|

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Latifah 9 months 2020-11-02T04:12:42+00:00 1 Answers 46 views 0

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    2020-11-02T04:13:42+00:00

    Giải thích các bước giải:

    Ta có:

    $|\vec{MA}+\vec{BC}|=|\vec{MC}+\vec{BC}|$
    $\to |\vec{MC}+\vec{CA}+\vec{BC}|=|\vec{MC}+\vec{BC}|$
    $\to |\vec{MC}+\vec{BA}|=|\vec{MC}+\vec{BC}|$

    $\to (|\vec{MC}+\vec{BA}|)^2=(|\vec{MC}+\vec{BC}|)^2$

    $\to (\vec{MC}+\vec{BA})^2=(\vec{MC}+\vec{BC})^2$

    $\to \vec{MC}^2+\vec{BA}^2+2\vec{MC}\cdot\vec{BA}=\vec{MC}^2+\vec{BC}^2+2\vec{MC}\cdot\vec{BC}$

    $\to MC^2+BA^2+2\vec{MC}\cdot\vec{BA}=MC^2+BC^2+2\vec{MC}\cdot\vec{BC}$

    $\to 2\vec{MC}\cdot\vec{BA}-2\vec{MC}\cdot\vec{BC}=BC^2-BA^2$

    $\to 2\vec{MC}(\vec{BA}-\vec{BC})=BC^2-BA^2$

    $\to 2\vec{MC}\cdot\vec{CA}=BC^2-BA^2$

    $\to 2\vec{MC}\cdot\vec{CA}\cdot\vec{CA}=(BC^2-BA^2)\vec{CA}$

    $\to 2\vec{MC}\cdot \vec{CA}^2=(BC^2-BA^2)\vec{CA}$

    $\to 2\vec{MC}\cdot CA^2=(BC^2-BA^2)\vec{CA}$

    $\to \vec{MC}=\dfrac{BC^2-AB^2}{2AC^2}\cdot \vec{CA}$

    $\to M\in AC,$ thỏa mãn $\vec{MC}=\dfrac{BC^2-AB^2}{2AC^2}\cdot \vec{CA}$

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