\(cos\left(B-C\right)=\frac{2.\left(2RsinB\right)\left(2RsinC\right)}{\left(2RsinA\right)^2}=\frac{2sinB.sinC}{sin^2A}\)
\(\Leftrightarrow sin^2A.cos\left(B-C\right)=2sinB.sinC=cos\left(B-C\right)-cos\left(B+C\right)\)
\(\Leftrightarrow cos\left(B-C\right)\left(1-sin^2A\right)=cos\left(B+C\right)=-cosA\)
\(\Leftrightarrow cos\left(B-C\right).cos^2A+cosA=0\)
\(\Leftrightarrow cosA\left(cos\left(B-C\right).cosA+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosA=0\Rightarrow A=90^0\\cos\left(B-C\right).cosA=-1\left(1\right)\end{matrix}\right.\)
Xét (1): do \(\left\{{}\begin{matrix}-1< cos\left(B-C\right)\le1\\-1< cosA< 1\end{matrix}\right.\)
\(\Rightarrow cos\left(B-C\right).cosA>-1\)
\(\Rightarrow\) Đẳng thức (1) không thể xảy ra
Vậy ABC vuông tại A
