Cho tam giác ABC có AB=c , BC=a , AC=b . Chứng minh Cos A = $\frac{b^{2}+c^{2}-a^{2} }{2bc}$

Kẻ $BH\bot AC$ $\Rightarrow \cos A=\dfrac{AH}{AB}\Rightarrow AH=AB.\cos A$

Theo định lý Pitago, ta có: $\begin{cases}BH^2=AB^2-AH^2\\BH^2=BC^2-CH^2\end{cases}$

$\Rightarrow AB^2-AH^2=BC^2-CH^2$

$\Rightarrow AB^2-AH^2+CH^2=BC^2$

$\Rightarrow AB^2-AH^2+\left( AC-AH \right)^2=BC^2$

$\Rightarrow AB^2-AH^2+AC^2-2AC.AH+AH^2=BC^2$

$\Rightarrow AB^2+AC^2-2AC.AH=BC^2$

$\Rightarrow AB^2+AC^2-2AB.AC.\cos A=BC^2$

$\Rightarrow \cos A=\dfrac{AB^2+AC^2-BC^2}{2AB.AC}$

$\Rightarrow \cos A=\dfrac{b^2+c^2-a^2}{2bc}$