Geometry problem in Circumcentre and Incentre

Question

I was trying to do this problem but could not. So I posted here for some hints or complete solution. if I is the incentre and S is the circumcenter of ABC prove that angle IAS is half the difference between angle B and angle C

Answer 1

I am using the notation you gave above..

Join $IA,IB,SA,SB,SC$

$\angle SBC = \angle SCB\ , \angle SAC=\angle SCA\ , \angle SAB=\angle SBA$

Equation 1: $\angle A/2+\angle IAS = \angle B/2+ \angle IBS$

$\implies 2\angle IAS-\angle B=2\angle IBS-\angle A$

Equation 2: $\angle C +\angle SBC +\angle SAC +\angle SBA + \angle SAB =180^0=\angle A+\angle B+\angle C $

$\implies 2\angle C +\angle IBS +\angle B/2+\angle IAS +\angle A/2= \angle A+\angle B+\angle C$

Solving these 2 equations, we get $2\angle IAS= \angle B-\angle C$