\(b\left( {{b^2} - {a^2}} \right) = c\left( {{a^2} - {c^2}} \right)\left( {a,b,c \ne 0} \right)\left( * \right)\)
Ta có: \(a^2=b^2+c^2-2bc.cosA\)
\(\Rightarrow\left\{{}\begin{matrix}a^2-c^2=b^2-2bc.cosA\\b^2-a^2=2bc.cosA-c^2\end{matrix}\right.\)
Thay vòa $(*)$ ta được:
\(\begin{array}{l} b\left( {2bc.\cos A - {c^2}} \right) = c\left( {{b^2} - 2bc.\cos A} \right)\\ \Leftrightarrow bc\left( {2b\cos A - c} \right) = bc\left( {b - 2c\cos A} \right)\\ \Leftrightarrow 2bc\cos A - c = b - 2c\cos A\left( {do:a,b,c \ne 0} \right)\\ \Leftrightarrow \cos A = \dfrac{1}{2} \Rightarrow \widehat A = {60^o} \end{array}\)
