Cho tam giac ABC. Tim M thoa : | vecto MA + vecto BC| = |vecto MC + vecto BC| 11/07/2021 Bởi Adalynn Cho tam giac ABC. Tim M thoa : | vecto MA + vecto BC| = |vecto MC + vecto BC|
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 $\vec{MC}=\dfrac{BC^2-AB^2}{2AC^2}\cdot \vec{CA}$ Bình luận
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 $\vec{MC}=\dfrac{BC^2-AB^2}{2AC^2}\cdot \vec{CA}$