Có $\frac{1}{sinA} + cotA=\frac{1+cosA}{sinA}=\frac{2cos^2\frac{A}2}{2cos\frac{A}2sin\frac{A}2}=\frac{cos\frac{A}2}{sin\frac{A}2}$ Mặt khác $\frac{a}{c-b}=\frac{sinA}{sinC-sinB}=\frac{sinA}{2sin\frac{C-B}2.sin\frac{A}2}=\frac{cos\frac{A}2}{sin\frac{C-B}2}$
$\Rightarrow sin\frac{A}2=sin\frac{C-B}2 \Leftrightarrow \widehat{C}=\widehat{A}+\widehat{B}$
$\Rightarrow \triangle ABC$ vuông ở C