Hint

Take $T$ to be the antipode of $A_1$, and apply Pascal on $C_1A_1A_1B_1TT$ and $C_1A_1B_1B_1TC_1$ to conclude.

Let $A_1B_1$ cut the parallel to $BC$ through $A$ at $L$, let $A'=AA'\cap (I)$, notice that $A'A_1B_1C_1$ is harmonic, then $A_1(K, L; A, B)=-1$ which is $A$ midpoint of $KL$, easy angle chase gives us $AK=AC_1$ and $AL=AB_1$ so we are done

We have $\angle AKC_1 = \angle C_1A_1B = \angle BC_1A_1 = \angle AC_1K$. Thus, $AK=AC_1=AB_1$. Then $\triangle AKB_1$ is isoceles, so $\angle AB_1K = \frac{1}{2} (180 - \angle KAB_1) = \frac{1}{2} \angle C = \angle ACI$. Thus $B_1K \parallel CI$. But $CI \perp A_1B_1$, so $KB_1 \perp A_1B_1$.

matinyousefi wrote: Let the incircle of triangle $ABC$ touches the sides $AB,BC,CA$ in $C_1,A_1,B_1$ respectively. If $A_1C_1$ cuts the parallel to $BC$ from $A$ at $K$ prove that $\angle KB_1A_1=90.$ $A$ is circumcenter of $\triangle KC_1B_1$

Let $\omega$ be a circle centered at $A$ with radius $AC_1=AB_1$. $$\angle AKC_1=\angle AKA_1=\angle BA_1K=\angle BA_1C_1=\angle A_1C_1B=\angle KC_1A$$Angle chasing above implies $AK=AC_1$ $\implies K\in\omega$ Now let $L$ be the intersection of $A_1B_1$ and $AK$. $$\angle ALB_1=\angle ALA_1=\angle CA_1L=\angle CA_1B_1=\angle A_1B_1C=\angle LB_1A$$From angle chasing above, $AL=AB_1 \implies L\in\omega$ So far we have a cyclic quadrilateral $KLB_1C_1$ centered at $A$ where $KL$ is diameter $\implies \angle KB_1A_1=KB_1L=90$

Since $AB_1=AC_1$ and, $\angle A_1KC_1=\angle BA_1C_1=\angle BC_1A_1=\angle AC_1K$ we have that the circumcenter of $(B_1C_1K)$ is $A$. So $\angle KB_1A_1=180-\angle B_1KC_1 - \angle C_1A_1B_1=180-\tfrac{1}{2}\angle A-\tfrac{1}{2}\angle B -\tfrac{1}{2}\angle C=180-90=90$

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/84.%201.%20Evaluation%20angles.pdf p. 8... Sincerely Jean-Louis