obviouly, IG∥AB is equal with a+b=2c so we need to prove CI⊥IO is equal with a+b=2c as attachment：

Attachments:

My complex bash solution: Let $A = a^2, B = b^2$ and $C = c^2$ be points on the unit circle in the complex plane. $ I = -ab-ac-bc, G = \frac{a^2+b^2+c^2}{3}, O = 0$ $\Rightarrow IG // AB \iff K_{IG} = K_{AB} = -a^2b^2 \iff \frac{\frac{a^2 + b^2 + c^2 + 3ab + 3ac +3bc}{3}}{\frac{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} + \frac{3}{ab} + \frac{3}{bc} + \frac{3}{ac}}{3}} = -a^2b^2 $ $\iff a^2 + b^2 + c^2 + 3ab + 3ac + 3bc = -b^2 - a^2 - \frac{a^2b^2}{c^2} - 3ab - \frac{3a^2b}{c} - \frac{3b^2a}{c} $ $\iff \frac{2a}{b} + \frac{2b}{a} + 6 + \frac{c^2}{ab} + \frac{ab}{c^2} + \frac{3b}{c} + \frac{3a}{c} + \frac{3c}{a} + \frac{3c}{b} = 0$, dividing both sides for $ab$. Note that $CI \perp IO \iff K_{CI} = -K_{IO} \iff \frac{c^2 + ab + ac + bc}{\frac{1}{c^2}+ \frac{1}{ab}+ \frac{1}{ac} + \frac{1}{bc}} = \frac{ab + ac + bc}{-\frac{1}{ab} - \frac{1}{ac} - \frac{1}{bc}}$ $\iff -\frac{c^2}{ab} - \frac{c}{a} - \frac{c}{b} - 1 - \frac{b}{c} - \frac{a}{c} - \frac{c}{b} - 1 - \frac{a}{b} - \frac{c}{a} - \frac{b}{a} - 1 = \frac{ab}{c^2} + \frac{a}{c} + \frac{b}{c} + 1 + \frac{c}{b} + \frac{c}{a} + \frac{b}{c} + 1 + \frac{b}{a} + \frac{a}{b} + 1$ $\iff \frac{ab}{c^2} + \frac{c^2}{ab} + \frac{3c}{a} + \frac{3c}{b} + \frac{3a}{c} + \frac{3b}{c} + \frac{2a}{b} + \frac{2b}{a} + 6 = 0$. So we get that $IG // AB \iff CI \perp IO$ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\blacksquare$

Let $D= CI\cap (ABC), E= CI\cap AB, M$ the midpoint of $BC$ and $N$ the midpoint of $AC.$ As $OA= OD$(ray)$\implies OAD$ isosceles. If $CI\perp OI$ $\implies I$ is the midpoint of $CD$. So , $IM//BD \implies \angle EDB= \angle CDB= \angle CIM$. Also, for $ABCD$ be cyclic, $\angle ICM= \angle DCB= \angle ACI= \angle EBD\implies \angle ICM= \angle EBD\implies CIM\sim BDE$. So, $\frac{CI}{BD}= \frac{CM}{BE}$. By the Incenter/Excenter Lemma, $BD= ID= DA\implies 1= \frac{CI}{ID}= \frac{CI}{BD}=\frac{CM}{BE} \implies CM= BE.$ Now, $2= \frac{BC}{CM}= \frac{BC}{EB}\implies \frac{BC}{EB}=2$. By the Angle Bisector Theorem, we know that $\frac{AC}{AE}=\frac{BC}{BE}= 2 \implies \frac{AC}{AE}=2$. On other hand, $N$ is the midpoint of $AC\implies \frac{AC}{AN}=2= \frac{AC}{AE}\implies CN=AN= AE$(1). Notice that $EAI\cong NAI(S.A.S)$, because $AI= AI, AE= AN$ and $\angle EAI=\angle NAI\implies IE= IN$. So, $NCI\cong EAD(S.A.S)$, because $DA=ID=CI\implies DA= CI, \angle NCI= \angle ACD= \angle DCB= \angle BAD=\angle EAD \implies \angle NCI= \angle EAD$ and we already had found that $AE= CN$, in (1). After that, we obtain $IN= ED$. So, $IE= IN= ED\implies \frac{CI}{IE}= \frac{CI}{\frac{ID}{2}}= 2$. Therefore, define $P$ the midpoint of $AB\implies \frac{CG}{GB}=\frac{CI}{IE}\implies IG//AB$ Now, what if $IG//AB$? So, we must have $CI\perp OI\Leftrightarrow IC= ID$. First, we see that, if $IG//AB\implies \frac{CI}{IE}=2$. By the Angle Bisector Theorem in $BCE$, we have $\frac{BC}{CI}= \frac{BE}{IE}\implies 2= \frac{CI}{IE}= \frac{BC}{BE}\implies \frac{BC}{BE}=2$. On the other hand, $\frac{BC}{BM}=2\implies BM=CM=BE$. Similarly, $AE= CN= AN$. Notice now that $EAI\cong NAI(S.A.S)\implies IN= IE= \frac{CI}{2}$. Let $I'$ the reflection of $I$ over $E$ in $CI$. So, $IE= EI'\implies II''= 2IE= CI\implies I$ is the midpoint of $CI'$, $NI//AD$ and as $N$ is the midpoint of $AC\implies NI= \frac{AD}{2}\implies AD= 2NI= 2IE= II'$. Similarly, $BD= II'\implies BD=AD= II'\Leftrightarrow I'$ is the midpoint of $\overset{\LARGE \frown}{\small{AB}}$, by the Incenter/Excenter Lemma $\implies I'= D\implies ID= IC$, as we wanted )

Attachments: