>>717
(1+cosA)/(b+c)=(sinB)/b=(sinC)/c=(sinB+sinC)/(b+c)
⇔1+cosA=sinB+sinC
⇔2{cos(A/2)}^2=2sin{(B+C)/2}cos{(B−C)/2}=2cos(A/2)cos{(B−C)/2}
⇔cos{(B−C)/2}=1
⇔B=C