$\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}=2R$hệ thức <=> $a=\frac{b+2c}{2cosB+cosC}$
=> $a(\frac{a^2+c^2-b^2}{ac}+\frac{a^2+b^2-c^2}{2ab})=b+2c$
=> $(a^2+c^2-b^2)2b+(a^2+b^2-c^2).c=(b+2c).2bc$
=> $2a^2b+2bc^2-2b^3+a^2c+b^2c-c^3-2b^2c-4bc^2=0$
=> $2a^2b-2bc^2-2b^3+a^2c-b^2c-c^3=0$
=> $(a^2-b^2-c^2)(2b+c)=0$