Din teorema sinusurilor obținem
[tex]\frac{c}{a+b}=\frac{\sin C}{\sin A+\sin B}=\frac{2\sin \frac C2 \cos \frac C2}{2\sin \frac{A+B}{2}\cos \frac{A-B}{2}}.[/tex]
Inlocuind in ipoteza, dupa simplificari, obtinem
[tex]\sin \frac{A+B}{2}\cos \frac{A-B}{2}=\cos \frac C2.[/tex]
Dar [tex]\sin \frac{A+B}{2}=\sin{\left(\frac{\pi}{2}-\frac C2}}\right)=\cos \frac C2,
[/tex]
de unde deducem [tex]\cos \frac{A-B}{2}=1,[/tex] adică [tex]A=B.[/tex]
