Xét $PT \Leftrightarrow \frac{sinA}{sinB}=\frac{2cos(\frac{B+C}2)cos(\frac{B-C}2)}{2cos(\frac{A+C}2)cos(\frac{A-C}2)}=\frac{sin\frac{A}2.cos\frac{B-C}2}{sin\frac{B}2.cos\frac{A-C}2}$$\Leftrightarrow \frac{cos\frac{A}2}{cos\frac{B}2}=\frac{cos\frac{B-C}2}{cos\frac{A-C}2}$
$\Leftrightarrow cos(\frac{2A-C}2)=coos(\frac{2B-C}2)$
$\Rightarrow \widehat{A}=\widehat{B}\Rightarrow \triangle ABC$ cân
Hoặc $\widehat{A}+\widehat{B}=\widehat{C} \Rightarrow \widehat{C}=90^0 \Rightarrow \triangle ABC vuông$