Question

Sa se arate ca un triunghi ABC in care are loc relatia sinC/2=c/(a+b) este isoscel. [ex. 312]

Answer 1

Din teorema sinusurilor obținem

$\frac{c}{a+b}=\frac{\sin C}{\sin A+\sin B}=\frac{2\sin \frac C2 \cos \frac C2}{2\sin \frac{A+B}{2}\cos \frac{A-B}{2}}.$

Inlocuind in ipoteza, dupa simplificari, obtinem

$\sin \frac{A+B}{2}\cos \frac{A-B}{2}=\cos \frac C2.$

Dar $\sin \frac{A+B}{2}=\sin{\left(\frac{\pi}{2}-\frac C2}}\right)=\cos \frac C2,$
de unde deducem $\cos \frac{A-B}{2}=1,$ adică $A=B.$