由正弦定理
a=2RsinA
b=2RsinB
1-cosA/(1-cosB)=a/b
1-cosA/(1-cosB)=sinA/sinB
(1-cosA)*sinB=(1-cosB)*sinA
(1-cosA)/sinA=(1-cosB)/sinB
提示tan(α/2)=sinα/(1+cosα)=(1-cosα)/sinα
故tan(A/2)=tan(B/2)
∴A=B
由正弦定理
a=2RsinA
b=2RsinB
1-cosA/(1-cosB)=a/b
1-cosA/(1-cosB)=sinA/sinB
(1-cosA)*sinB=(1-cosB)*sinA
(1-cosA)/sinA=(1-cosB)/sinB
提示tan(α/2)=sinα/(1+cosα)=(1-cosα)/sinα
故tan(A/2)=tan(B/2)
∴A=B