Let $D$ and $E$ be the midpoints of minor arcs $AC$ and $BC$, respectively; Pascal on $BDEACC$ implies that $X\equiv DE\cap CC$ lies on $A_1B_1$. Then since $DE$ is the perpendicular bisector of segment $\overline{CI}$, we have that $XC=XI$, so $\angle CIX=\angle XCI=\angle B + \tfrac{\pi}{4}$, and so $IX\parallel AB$. Let $H$ be the foot of the altitude from $C$; it is easy to see that $CX$ passes through the harmonic conjugate of $H$ with respect to $\overline{AB}$, so projecting through $C$ onto $A_1B_1$ and then through $I$ onto $AB$ implies the conclusion.

Another vectorial approach: Denote the foot of the altitude $D$, and the foot of the angle bisector $T$. Denote $X$ the intersection of $CD$ and $A_1B_1$. Then by Transversales theorem we get $\frac{DX}{XC}=\frac{x+y}{z}$, and $\frac{AI}{IT}=\frac{x+y}{z}$. We want to prove that $X-I-C_O$ are collinear, equivalent with $XC_0=n*IC_O$ (vectors), but $XC_O=\frac{AX+BX}{2}$ and $IC_O=\frac{AI+BI}{2}$ Now substituting $AX=\frac{AD+AC*\frac{x+y}{z}}{\frac{x+y}{z}+1}$, same for $BX$ and $AI=\frac{AT+AC*\frac{z}{x+y}}{\frac{z}{x+y}+1}$ (same for $BI$) we'll get the desired result.

$\frac{\sin \angle AIC_0}{\sin \angle C_0IB }=\frac{IB}{IA},\frac{\sin \angle A_1IC}{\sin \angle CIB_1 }\cdot \frac{IA_1}{IB_1}= \frac{KA_1}{KB_1}$ where $K\equiv A_1B_1\cap IC_0$ hence $ \frac{KA_1}{KB_1}=\frac{IA_1}{IB_1}\cdot \frac{IB}{IA}=\frac{A_1C}{AC}\cdot \frac{BC}{B_1C}=\frac{A_1C}{B_1C}\cdot\frac{\sin \angle A}{\sin \angle B}$ besides $\frac{K'A_1}{K'B_1}=\frac{CA_1}{CB_1}\cdot \frac{\sin angle A}{\sin \angle B} $ where $K'\equiv A_1B_1\cap AH$ ($AH$the altitude ). Therefore $K'=K$ RH HAS

Dear Mathlinkers, see also http://jl.ayme.pagesperso-orange.fr/Docs/La%20ponctuelle%20(MI).pdf p. 16-17. Sincerely Jean-Louis

See the proposed problem P7 from here => https://artofproblemsolving.com/community/c1090h1313367

Denote by : $C_1=AB \cap CI$ ; $C_2=AB \cap A_1B_1$ ; $D=C_0I \cap A_1B_1$ ; $S=CI \cap A_1B_1$ ; $T=CC_0 \cap A_1B_1$ ; $U=C_2I \cap BC$ . In triangle $\triangle C_2BA_1$ , $BB_1$ and $AA_1$ are cevians , therefore $(C,U,A_1,B) = -1$ and pencil $(C_2,C;U,A_1,B)$ is harmonic. Thus based on Pappus' Theorem any line passing through this pencil generates a harmonic ratio. If we look at $CC_1$ we get $(C,I,S,C_1)=-1$. Then pencil $(C_0;C,I,S,C_1)$ is also harmonic and based on the same Pappus theorem its intersection with $A_1B_1$ also generates a harmonic ratio. This means that $(C_2,S,D,T)=-1$ $(1)$. On the other hand $BB_1$ and $AA_1$ are also cevians in triangle $\triangle ABC$ , therefore $(C_2,C_1,A,B)=-1$ and because $CC_1$ is the angle bisector of $\angle ACB$ we get that $CC_2$ is the external bisector of $\angle ACB$. Then $\angle SCC_2=90^0$ . $(2)$. From $(1)$ and $(2)$ we get that $CS$ is the angle bisector of $\angle DCT$ , therefore $DC$ and $DT$ are isogonal in $\triangle ABC$. $(3)$ But $C_0$ is a median in right triangle $\triangle ABC$ which means that $C_0$ is the circumcenter of triangle $\triangle ABC$ . $(4)$ From $(3)$ and $(4)$ the result follows because in any triangle the altitude from any vertex and the segment joining that vertex with the circumcenter are isogonal.

Let the intersection of the altitude from $C$ and $A_1B_1$ be $P$. Let $P_\infty$ be the point at infinity along line $AB$. Projecting $-1=(A,B;C_0P_\infty)$ at $I$ onto $A_1B_1$, we want $(A_1,B_1;P,A_1B_1\cap \ell)=-1$, where $\ell$ is the line through $I$ parallel to $BC$. (This is implicitly phantom points.) Call $A_1B_1\cap \ell$ as $X$. Note that the altitude from $C$ is actually the $C$-symmedian. Projecting at $C$, we want $CX$ to be the tangent to $(ABC)$ at $C$. Take a homothety at $C$ sending $\ell$ to $AB$. The $C$-tangent remains fixed, $\ell$ is sent to $AB$ (by definition). Note that the ratio of the homothety is $CD/CI$, where $D$ is the intersection of the $C$-bisector and $AB$. Using mass points (!) or any other method, this ratio is found to be $\frac{a+b+c}{a+b}$. By angle bisector theorem, $CB_1=b\cdot \frac{a}{a+c}$ and $CA_1=a\cdot \frac{b}{b+c}$. So $CB_1'=\frac{a+b+c}{a+b}\cdot \frac{ab}{a+c}=\frac{b(a^2+ab+ac)}{a^2+ab+ac+bc}=b\cdot \left(1-\frac{bc}{(a+b)(a+c)}\right)$. It follows that $B_1'A=b\cdot\frac{bc}{(a+b)(a+c)}$. Let the image of $X$ be $X'$. Let $X'^*$ be the intersection of the tangent at $C$ with $AB$. We wish to show that $AX'/X'B=b^2/a^2$, implying $X'\equiv X'^*$. By Menelaus, we need \[\frac{BX'}{X'A}\cdot\frac{AB_1'}{B_1'C}\cdot\frac{CA_1'}{A_1'B}=1.\]The second ratio equals $\frac{\frac{bc}{(a+b)(a+c)}}{1-\frac{bc}{(a+b)(a+c)}}$ and similarly the third equals $\frac{1-\frac{ac}{(b+a)(b+c)}}{\frac{ac}{(b+a)(b+c)}}$. Multiplying them, we get \[\frac{\frac{bc}{(a+b)(a+c)}}{1-\frac{bc}{(a+b)(a+c)}}\cdot \frac{1-\frac{ac}{(b+a)(b+c)}}{\frac{ac}{(b+a)(b+c)}}=\frac{bc}{a^2+ab+ac}\cdot \frac{b^2+ab+bc}{ac}=\frac{b^2(a+b+c)}{a^2(a+b+c)}=\frac{b^2}{a^2},\]as desired.