Let $L \equiv AI \cap BC,$ $D \equiv MN \cap BC,$ $Y \equiv AM \cap BC$ and $U \equiv BP \cap AI.$ As $MN$ is the perpendicular bisector of $AI$ (well-known), then $K \in MN.$ Since $BU,BL$ are the reflections of $BK,BA$ across $BI,$ we deduce that $(L,I,A,K)=(A,K,L,I)=(L,U,A,I)=B(L,U,A,I)=(Y,P,A,M)$ $\Longrightarrow$ $D(L,I,A,K)=D(Y,P,A,M)$ $\Longrightarrow$ $P \in DI.$ By similar reasoning $Q \in DI$ $\Longrightarrow$ $P,Q,I$ are collinear.

I have seen general problem Let $ABC$ be a triangle inscribed in circle $(O)$ and $P,Q$ are two isogonal conjugate points. $PB,PC$ cut $(O)$ again at $M,N$. $QA$ cuts $MN$ at $K$. $L$ is isogonal conjugate of $K$. $LB,LC$ cut $AM,AN$ at $S,T$, resp. Prove that $S,Q,T$ are collinear.

buratinogigle wrote: I have seen general problem Let $ABC$ be a triangle inscribed in circle $(O)$ and $P,Q$ are two isogonal conjugate points. $PB,PC$ cut $(O)$ again at $M,N.$ $QA$ cuts $MN$ at $K.$ $L$ is isogonal conjugate of $K.$ $LB,LC$ cut $AM,AN$ at $S,T,$ resp. Prove that $S,Q,T$ are collinear. The proof to this generalization is very similar to what I did in my previous solution. Letting $D \equiv MN \cap BC,$ $Y \equiv AM \cap BC,$ $U \equiv AQ \cap BC$ and $ V \equiv AP \cap BC,$ we get $B(Y,S,A,M)=B(V,L,A,P)=(A,K,U,Q)=(U,Q,A,K)$ $\Longrightarrow$ $D(Y,S,A,M)=D(U,Q,A,K)$ $\Longrightarrow$ $S \in DQ$ and likewise $T \in DQ$ $\Longrightarrow$ $S,Q,T$ are collinear.

Hello Please give some other solutions to see different ideas I have got that $BP,CQ$ interest each other on $AI$ At a point like $S$ and we can easily get that $(I_{a}ISK)=-1$ Can this be useful?!

we'll solve the problem using bary coordinates. let $\triangle ABC$ be the refrence triangle and let $k'=(QC,BP)$ and literally $K,K'$ are isogonal conjugates and we have that: $K=(2a+b+c:b:c) \implies K'=(\frac{a^2}{2a+b+c}:b:c)$ and now we can easily calculate $Q$ from cevians $CK',AN$ and already know that: $N=(a:b:\frac{-c^2}{a+b}), M=(a:\frac{-b^2}{a+c}:c) \implies Q=(\frac{a^2}{2a+b+c}:b:\frac{-c^2}{a+b}), P=(\frac{a^2}{2a+b+c}:\frac{-b^2}{a+c}:c)$ and it's pretty much straightforward to show $\begin{vmatrix} \frac{a^2}{2a+b+c} & b &\frac{-c^2}{a+b} \\ a & b &c \\ \frac{a^2}{2a+b+c} & \frac{-b^2}{a+c} & c \\ \end{vmatrix}=0$

Iran MO Round 3 2015 G4 wrote: Let $ABC$ be a triangle with incenter $I$. Let $K$ be the midpoint of $AI$ and $BI\cap \odot(\triangle ABC)=M,CI\cap \odot(\triangle ABC)=N$. points $P,Q$ lie on $AM,AN$ respectively such that $\angle ABK=\angle PBC,\angle ACK=\angle QCB$. Prove that $P,Q,I$ are collinear. Solution (with Aryan-23): Let $MN \cap BC=S$, $AN \cap BC=E$ & $AI$ $\cap$ $BC$ $=$ $U$. Let $K'$ be the isogonal conjugate of $K$ WRT $\Delta ABC$ $$(A,Q;N,E) \overset{C}{=} (A,K';I,U) =(U,K;I,A)=(A,I;K,U)$$Hence, $S,P,I$ collinear and similarly, $Q,I,S$ collinear $\qquad \blacksquare$

Let $B' , C'$ be points on $BK , CK$ such that $AB'||BI$ and $AC'||CI$ , so $CK=CK'$ and $BK=BK'$ then $BC'||CB'$ and we can get $\angle{ABC'}+\angle{ACB'}=\angle{BAC}$ $(1)$ It's easy to check that $P , B'$ are isogonal conjucate and $Q , C'$ are , too . If $ S $ is intersection of $QB , CP$ , we can check easily from $(1)$ that $ S $ is on the circle of $ABC$ . So with pascal on $NCSBMA$ we can get $I , P , Q $ are collinear $\blacksquare$

isogonality lemma kills it Let $MN \cap BC =R$ note that $(CK,CQ) , (CA,CR)$ are pairs of isogonal line wrt $\angle BCA$ since $N=AQ \cap KR $ we have that $RQ \cap AK$ is isogonal to $AN$ so $RQ \cap AK ,A,N,I$ are collinear so $I=RQ \cap AK$ similarly $I=PQ \cap AK$