If I is the incenter of the circle inscribed in the triangle ABC and AI cuts BC at P, then

Explanation

`(AP)/(SinC)=b/(sin(B+A/2))`

`(AP)=(bSinC)/(Sin(B+A/2)`

`AI=r/(SinA/2)`

`r cotA/2+rcotB/2=C`

`r=C/(cotA/2+CotB/2)`

`r=(CSina/2*Sinb/2)/(Cos(C/2)`

We need to find the ratio of AI and AP

`(AI)/(AP)=(C*sinB/2)/(CosC/2)*Sin(B+A/2)/(bSinC)`

...