AB=\(\sqrt{\left[4-\left(-3\right)\right]^2+\left(-1-2\right)^2}=\sqrt{58}\)

\(AC=\sqrt{\left(4-1\right)^2+\left(-1-6\right)^2}=\sqrt{58}\)

=>tam giác ABC cân tại A

\(BC=\sqrt{\left(-3-1\right)^2+\left(2-6\right)^2}=4\sqrt{2}\)

=>BC/2=\(2\sqrt{2}\)

Suy ra: \(\sin\frac{1}{2}BAC=\frac{\frac{BC}{2}}{AC}=\frac{2\sqrt{2}}{\sqrt{58}}\Rightarrow\frac{1}{2}\text{góc BAC}\approx22^0\Rightarrow\text{góc BAC}\approx11^0\)

nhầm nha 

=>góc BAC=44 độ

\(AB=\sqrt{\left(-3-4\right)^2+\left(2+1\right)^2}=\sqrt{58}\)

\(AC=\sqrt{\left(1-4\right)^2+\left(6+1\right)^2}=\sqrt{58}\)

\(BC=\sqrt{\left(1+3\right)^2+\left(6-2\right)^2}=4\sqrt{2}\)

\(\cos BAC=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\dfrac{58+58-32}{2\cdot58}=\dfrac{84}{116}\)

nên \(\widehat{BAC}\simeq44^0\)

\(\overrightarrow{AB}=\left(4;4\right)\)

Gọi \(D\left(x;y\right)\) \(\Rightarrow\overrightarrow{DC}=\left(4-x;-1-y\right)\)

ABCD là hbh \(\Leftrightarrow\overrightarrow{AB}=\overrightarrow{DC}\)

\(\Leftrightarrow\left\{{}\begin{matrix}4-x=4\\-1-y=4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0\\y=-5\end{matrix}\right.\) \(\Rightarrow D\left(0;-5\right)\)

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