Quote: Let $\omega$ denote the circumscribed circle of triangle $ABC$, $I$ be its incenter, and $K$ be any point on arc $AC$ of $\omega$ not containing $B$. Point $P$ is symmetric to $I$ with respect to point $K$. Point $T$ on arc $AC$ of $\omega$ containing point $B$ is such that $\angle KCT = \angle PCI$. Show that the bisectors of angles $AKC$ and $ATC$ meet on line $CI$. Let $D$ be intersection of $\overline{CI}$ and the angle bisector of $\angle AKC$. Let $I_B$ be the $B$-excenter of $\triangle ABC$ and let $M$ be the midpoint of arc $AC$, not containing $B$. Let $Q$ be reflection of $I$ over $D$. Claim. $I_BPQC$ is cyclic. Proof. By homothety, it is sufficient to show that $\measuredangle MKD=90^\circ$. Firstly, observe that $AIKD$ is cyclic since $\measuredangle DIA=90^\circ+\tfrac{1}{2}\measuredangle CBA=\tfrac{1}{2}\measuredangle CKA=\measuredangle DKA$. Now indeed, \begin{align*} \measuredangle MKD=\measuredangle MKA+\measuredangle AKD=\measuredangle IBA+\measuredangle AID=90^\circ, \end{align*}as desired. $\square$ Now, \begin{align*} \measuredangle DMK=\measuredangle QI_BP=\measuredangle QCP=\measuredangle ICP=\measuredangle TCK=\measuredangle TMK, \end{align*}which means that $\overline{TM}$ passes through $D$, we are done. $\blacksquare$

rafaello wrote: Am I tired or something but this problem also does not make any sense for me, could you check your problem. Sorry, I checked and everything seems right, what exactly is your problem?

MS_Kekas wrote: Let $\omega$ denote the circumscribed circle of triangle $ABC$, $I$ be its incenter, and $K$ be any point on arc $AC$ of $\omega$ not containing $B$. Point $P$ is symmetric to $I$ with respect to point $K$. Point $T$ on arc $AC$ of $\omega$ containing point $B$ is such that $\angle KCT = \angle PCI$. Show that the bisectors of angles $AKT$ and $ATC$ meet on line $CI$. (Proposed by Anton Trygub) Should it be $AKC$ and $ATC$ in the last line?

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Fedor Bakharev wrote: MS_Kekas wrote: Let $\omega$ denote the circumscribed circle of triangle $ABC$, $I$ be its incenter, and $K$ be any point on arc $AC$ of $\omega$ not containing $B$. Point $P$ is symmetric to $I$ with respect to point $K$. Point $T$ on arc $AC$ of $\omega$ containing point $B$ is such that $\angle KCT = \angle PCI$. Show that the bisectors of angles $AKT$ and $ATC$ meet on line $CI$. (Proposed by Anton Trygub) Should it be $AKC$ and $ATC$ in the last line? Oh thanks, my bad. Corrected!