Let $D=BI \cap (ABC)$. Now apply ptolemy theorem $$AB.DC+BC.AD=AC.BD$$ which is simply to $BD = 2DA$ and we know $AD=DI=DC$ which give us $I$ is midpoint of $BD$ so $\angle OIB =90$

Same as Bulgarian Autumn 2022 9.2 Put $M$ and $N$ as midpoints of $AB$ and $BC$ and $K$ and $L$ as feet from $I$ to $AB$ and $BC$, check that $2b = a+c$ implies $KM = LN$ and congruent triangles imply that $B$, $M$, $N$, $I$, $O$ are concyclic, done!

Consider R and r equal to the length of the radius of the circumcircle and incircle . Lemma1 : $ AI = (bc/p).cosA/2 $ . Lemma2 : $ S = 1/2.ac.sinB = abc/4R $ . Lemma3 : $ r = s/p $. According to the Pythagorean theorem, it is enough to prove $ OI^2 + BI^2 = R^2 $. According to the Euler's relation,$ OI^2 = R^2 -2Rr $. It is enough to prove $ BI^2 = 2Rr $. $ 2Rr = 2R.(s/p) = 2R.( (1/2.ac.sinB)/p ) = ( (ac/p)cosB/2 )^2 \Longleftrightarrow abc/4R = (b.sinB/2) / (2.cosB/2) \Longleftrightarrow S = (b.sinB/2) / (2.cosB/2) $. Can anyone complete this solution?