There's a typo:N lie on BC my solution: Let U,V,W be the midpoint of MN,AC,PQ It's sufficient to prove U,V,W,C are concyclic which is equivaient to KV+KU=$\frac{1}{2}(KM+KA-MA+KN+KC+NC)=\frac{1}{2}(KM+KN+KA+KC)$

Denote $\left( {KAM} \right) \cap KP = U$,$\left( {KCN} \right) \cap KQ = V$. Then $\angle UAM = \angle UMA = \frac{1}{2}\angle K = \angle VCN = \angle VNC$，so $\Delta UAM \cong \Delta VCN$，and we have $UP = UA = VC = VQ$, so $RP = RQ \Leftrightarrow RU = RV$. But from $\Delta RAM \cong \Delta RCN$，we have $\Delta RUM \cong \Delta RVN$, so we done.

Attachments:

$R$ is the Miquel point of the quadrilateral $AMNC$, ie there is a rotation centered in $R$ which sends $\odot(DNC)$ to $\odot(DMA)$, therefore the image of the midpoint of the small arc $(NC)$ will be the midpoint of the small arc $(MA)$, and this immediately implies the problem. See here: http://www.cut-the-knot.org/Curriculum/Geometry/SpiralSim.shtml

Note that $AR=CR, AM=CN, \angle RAM=\angle RCN$, so $RAM\cong RCN$. So $R$ is the miquel point of complete quadrilateral $BMACNK$. Then $R$ lies on the perpendicular bisectors of $MN, AC$. Let $D,E$ be the midpoints of $MN,AC$. Suppose $(KDER)$ meets $KPQ$ at $X$; it suffices to show $X$ is the midpoint of $PQ$. Let $(KAM),(KNC)$ meet $KPQ$ at $F,G$ respectively. Note that $X$ is the midpoint of $FG$. Now $F$ is the arc midpoint of $AM$ in $(KAM)$, so it is the center of circle $PAM$. Thus \[ FP=FM=\frac{AM}{2\cos \angle AKM/2}=\frac{CN}{2\cos \angle CKN/2}=GN=GQ \] so $X$ is the midpoint of $PQ$.

Let $I$ be the intersection of $KP$ with the circimcircle of $KAM$ and $J$ be the intersection point of $KQ$ with the circumcircle of $KCN$.Now,because $AMI$ is congruent to $CNJ$ we have $IP = JQ$ so it is enough to prove $RI=RJ$.It is easy to show that $RAM$ and $RNC$ are congruent so from $MI=NJ$ and the previous congruence we have $RMI$ is congruent to $RNJ$ so we are finished.

junioragd wrote: Let $I$ be the intersection of $KP$ with the circimcircle of $KAM$ and $J$ be the intersection point of $KQ$ with the circumcircle of $KCN$.Now,because $AMI$ is congruent to $CNJ$ we have $IP = JQ$ so it is enough to prove $RI=RJ$.It is easy to show that $RAM$ and $RNC$ are congruent so from $MI=NJ$ and the previous congruence we have $RMI$ is congruent to $RNJ$ so we are finished. can you explain why RAM and RNC are congruent

yunxiu wrote: Denote $\left( {KAM} \right) \cap KP = U$,$\left( {KCN} \right) \cap KQ = V$. Then $\angle UAM = \angle UMA = \frac{1}{2}\angle K = \angle VCN = \angle VNC$，so $\Delta UAM \cong \Delta VCN$，and we have $UP = UA = VC = VQ$, so $RP = RQ \Leftrightarrow RU = RV$. But from $\Delta RAM \cong \Delta RCN$，we have $\Delta RUM \cong \Delta RVN$, so we done. how can you prove that RAM and RCN are congruent

[asy][asy] unitsize(1.5inches); pair B=dir(125); pair A=dir(215); pair C=dir(-35); pair I=incenter(A,B,C); pair L=circumcenter(A,I,C); pair R=-L; pair M=0.73*B+(1-0.73)*A; pair N=2*foot(circumcenter(B,R,M),B,C)-B; pair K=extension(M,N,A,C); pair P=incenter(A,M,K); pair Q=extension(P,K,C,I); pair PP=((C-R)/(A-R))*(P-R)+R; pair QQ=((A-R)/(C-R))*(Q-R)+R; draw(A--B--C--cycle); draw(circumcircle(A,B,C)); draw(circumcircle(B,M,N)); draw(C--K); draw(M--K); draw(circle(R,abs(R-P)),dotted); draw(QQ--M--P--A--cycle); draw(PP--N--Q--C--cycle); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(-60)); dot("$R$",R,dir(R)); dot("$M$",M,dir(30)); dot("$N$",N,1.5*dir(80)); dot("$K$",K,dir(0)); dot("$P$",P,dir(-75)); dot("$Q$",Q,dir(-110)); dot("$P'$",PP,dir(0)); dot("$Q'$",QQ,dir(70)); [/asy][/asy] Claim: $R\in(BMN)$. Proof: Let $R'=(BMN)\cap(BAC)$. We have a spiral similarity $\psi$ centered at $R'$ sending $MA\mapsto NC$. Since $MA=NC$, this has to be a rotation, so $R'A=R'C$, so $R'=R$, as desired. We also see that $RM=RN$ since $\psi(M)=N$. $\blacksquare$ Let $P'=\psi(P)$ and $Q'=\psi^{-1}(Q)$. Since $Q$ is a rotation, we see that \[P'NQC=\psi(PMQA),\]so $P'NQC\cong PMQ'A$. Note that $\angle MPA=\pi/2+\frac{1}{2}K$ and $\angle NQC=\pi/2-\frac{1}{2}K$, so $\angle MPA+\angle NQC=\pi$. Thus, $\angle NP'C+\angle NQC=\pi$, so actually the two congruent quadrilaterals $P'NQC$ and $PMQ'A$ are cyclic. We claim the quadrilateral $P'QPQ'$ is cyclic. Since $Q'P=PQ'$, it suffices to show that it is an isosceles trapezoid, or that $\angle P'QP=\angle Q'PQ$. We see that \begin{align*} \angle Q'PQ &= \angle Q'PM+\angle +\angle MPQ \\ &=\angle Q'AM+\angle MPK \\ &=\angle QCN+\angle (\pi/2+A/2) \\ &=\pi/2+A/2+C/2, \end{align*}and by symmetry then, we also have $\angle P'QP=\pi/2+A/2+C/2$, so $P'QPQ'$ is cyclic, as desired. Since $RP=RP'$ and $RQ=RQ'$, we see that the circumcenter of $(P'QPQ')$ is $R$, so $RP=RQ$, as desired.

ARMO 2014 Grade 10 P4 wrote: Given a triangle $ABC$ with $AB>BC$, let $ \Omega $ be the circumcircle. Let $M$, $N$ lie on the sides $AB$, $BC$ respectively, such that $AM=CN$. Let $K$ be the intersection of $MN$ and $AC$. Let $P$ be the incentre of the triangle $AMK$ and $Q$ be the $K$-excentre of the triangle $CNK$. If $R$ is midpoint of the arc $ABC$ of $ \Omega $ then prove that $RP=RQ$. Claim 1:- $\triangle RMA\cong\triangle RNC$. Easy to see that $RC=RA$, $MA=NC$ and $\angle NCR=\angle MAR$. So, $MR=RN$, After a bit of angle chasing we find that $RBNM$ is a cyclic quadrilateral. Hence, $R$ is the Miquel Point of $CMNA$. Hence, $R\in$ {$\odot(KAM),\odot(KCN)$}. Now notice that as triangles $RMA,RNC$ were congruent, their Circumcircles will also be congruent. Let $KP\cap\odot(KAM)=D$ and $KP\cap\odot(KCN)=E$. Claim 2:- $\triangle REC\cong\triangle RDA$. $RC=RA$. $\angle ECR=\angle EKR=\angle DKR=\angle DAR$. Now notice that $\frac{DA}{\sin DKA}=\frac{EC}{\sin DKA}\implies DA=EC$. Note that we took this step just because their Circumcircles are congruent. Hence, $RE=RP$. Claim 3:- $\triangle REQ\cong\triangle RDP$. $RE=RP$, $\angle RFQ=\angle RDP$ and $QE=EC=DA=DP$ by IEL or Fact $5$. Hence, $\boxed{RP=RQ}$.

Define $X, Y$ as the midpoints of minor $\widehat{AM}$ and $\widehat{CN}$ respectively. First we claim that $R$ is the Miquel Point of $AMNC$. Indeed, the Miquel Point $R'$ satisfies $R'A=R'C$, and $R'$ also lies on $(ABC)$ by definition. Thus $R=R'$. Now $RM=RN$ by spiral similarity, and now $$\measuredangle RXK = \measuredangle RNK = \measuredangle N MR = \measuredangle KMR = \measuredangle KYR = \measuredangle XYR,$$so $RX=RY$. Next, the circumcircles of $(KAM)$ and $(KCN)$ are congruent, whilst the distance between incenter and excenter formula implies $PP_1 = QQ_1$, where $P_1$ is the $K$-excenter of $\triangle KAM$ and $Q_1$ is the incenter of $\triangle KCN$. Now $P_1$ is the reflection of $P$ over $Y$ and similarly for $Q_1$, and thus we are done.

Let $R'$ be the Miquel Point of $AMNY$. Because $AM = NY$, $R$ lies above $BC$, on the perpendicular bisector of $AB$, and on $(ABC)$, so $R = R'$. Next, let $X$ and $Y$ be the intersection of $PQ$ with $(KAMR)$ and $(KCNR)$ where $X, Y \ne K$. Notice that the spiral similarity at $R$ brings $AXM$ to $CYN$ so $RX = RY$. Also by Incenter-Excenter, we have $PX = AX = CY = YQ$ and clearly $\angle{RXP} = \angle{RYQ}$ so $RP = RQ$ by SAS Congruency, as desired. $\blacksquare$

Note that $R$ is the Miquel point of the complete quadrilateral $NMAC$. Moreover, since $NC = AM$, the spiral similarity from $\overline{NC}$ to $\overline{MA}$ actually gives rise to a congruence $\triangle RNC \cong \triangle RMA$. [asy][asy] size(12cm); pair B = dir(130); pair C = dir(220); pair A = dir(320); draw(A--B--C--cycle); pair R = dir(90); draw(unitcircle); pair M = 0.8*dir(B-A)+A; pair N = 0.8*dir(B-C)+C; pair K = extension(M, N, A, C); pair P = incenter(A, M, K); draw(N--K--C); draw(circumcircle(K, N, C), gray); draw(circumcircle(K, M, A), gray); pair X = circumcenter(P, A, M); pair Y = -X+2*foot(R, P, K); pair Q = -P+2*foot(R, P, K); draw(K--Q); draw(circumcircle(B, N, M), lightblue); draw(X--R--Y, red); draw(Q--R--P, dashed+red); clip(scale(2.6)*unitcircle); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$A$", A, dir(280)); dot("$R$", R, dir(R)); dot("$M$", M, dir(85)); dot("$N$", N, dir(45)); dot("$K$", K, dir(K)); dot("$P$", P, dir(80)); dot("$X$", X, dir(-90)); dot("$Y$", Y, dir(115)); dot("$Q$", Q, dir(Q)); /* Source generated by TSQ */ /* B = dir 130 C = dir 220 A = dir 320 A--B--C--cycle R = dir 90 unitcircle M = 0.8*dir(B-A)+A R45 N = 0.8*dir(B-C)+C R45 K = extension M N A C P = incenter A M K N--K--C circumcircle K N C dotted circumcircle K M A dotted X = circumcenter P A M R-90 Y = -X+2*foot R P K R135 Q = -P+2*foot R P K K--Q circle B N M dotted blue X--R--Y red Q--R--P dotted red !clip(scale(2.6)*unitcircle); */ [/asy][/asy] Let $X$ and $Y$ be the arc midpoints of $\widehat{MA}$ and $\widehat{NC}$, that is, the circumcenters of $\triangle MPA$ and $\triangle NQC$ (by the incenter/excenter lemma). Since $\triangle RNC \cong \triangle RMA$, it follows $RX = RY$ too. As $QY = XP$ we then deduce $RP = RQ$, as required.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -16.967493731979896, xmax = 26.352373619191397, ymin = -11.818498683493507, ymax = 15.844786764876519; /* image dimensions */ pen ccwwqq = rgb(0.8,0.4,0); pen ffqqtt = rgb(1,0,0.2); pen ccqqqq = rgb(0.8,0,0); pen cczzqq = rgb(0.8,0.6,0); pen ffwwqq = rgb(1,0.4,0); pen ttqqqq = rgb(0.2,0,0); /* draw figures */ draw((-9.518726518817983,-4.4884190523946765)--(2.1034937071263995,-4.4884190523946765), linewidth(0.9) + ccwwqq); draw((2.1034937071263995,-4.4884190523946765)--(-8.537774675900362,8.262820666697259), linewidth(0.9) + ccwwqq); draw(circle((-3.7076164058457923,1.4778848480260836), 8.328612308017684), linewidth(0.9) + ffqqtt); draw((-9.08151781712523,1.194788541135947)--(-1.5486423287168227,-0.1121303992440007), linewidth(0.9) + ccwwqq); draw((-8.537774675900362,8.262820666697259)--(-9.518726518817983,-4.4884190523946765), linewidth(0.9) + ccwwqq); draw((2.1034937071263995,-4.4884190523946765)--(-9.08151781712523,1.194788541135947), linewidth(0.9)); draw((-1.5486423287168227,-0.1121303992440007)--(-9.518726518817983,-4.4884190523946765), linewidth(0.9)); draw((-1.5486423287168227,-0.1121303992440007)--(23.675598063678652,-4.488419052394677), linewidth(0.9) + ccwwqq); draw((2.1034937071263995,-4.4884190523946765)--(23.675598063678652,-4.488419052394677), linewidth(0.9) + ccwwqq); draw((23.675598063678652,-4.488419052394677)--(-12.394797136510528,-1.3825945103388317), linewidth(0.9) + ccqqqq); draw(circle((12.889545885402526,8.224894500337271), 16.672350256714648), linewidth(0.9) + cczzqq); draw(circle((7.078435772430335,-2.906816396688174), 16.672350256714648), linewidth(0.9) + ffwwqq); draw(circle((-12.394797136510528,-1.3825945103388317), 3.1058245420558426), linewidth(0.9) + dotted + ccqqqq); draw((-11.863882189993353,1.6775158480634738)--(-9.08151781712523,1.194788541135947), linewidth(0.9) + dotted + ttqqqq); draw((-9.518726518817983,-4.4884190523946765)--(-12.394797136510537,-4.488419052394691), linewidth(0.9) + dotted); draw(circle((2.939016543329883,-2.702905379844089), 1.7855136725505885), linewidth(0.9) + dotted + ccqqqq); /* dots and labels */ dot((-8.537774675900362,8.262820666697259),dotstyle); label("$A$", (-8.950281324667478,8.685196043572596), NE * labelscalefactor); dot((-9.518726518817983,-4.4884190523946765),dotstyle); label("$B$", (-10.447239291663067,-5.697691148825591), NE * labelscalefactor); dot((2.1034937071263995,-4.4884190523946765),dotstyle); label("$C$", (1.4400726809622726,-5.717635839127228), NE * labelscalefactor); dot((-9.08151781712523,1.194788541135947),linewidth(4pt) + dotstyle); label("$M$", (-10.398124246718896,1.6447203670948616), NE * labelscalefactor); dot((-1.5486423287168227,-0.1121303992440007),linewidth(4pt) + dotstyle); label("$N$", (-1.4704693618785618,0.04859204598029653), NE * labelscalefactor); dot((-12.394797136510528,-1.3825945103388317),linewidth(4pt) + dotstyle); label("$Q$", (-13.018998147705665,-1.2373703460391784), NE * labelscalefactor); dot((23.675598063678652,-4.488419052394677),linewidth(4pt) + dotstyle); label("$K$", (23.85928733056506,-4.997967697317408), NE * labelscalefactor); dot((2.939016543329883,-2.702905379844089),linewidth(4pt) + dotstyle); label("$P$", (2.517085957648852,-2.244217693232467), NE * labelscalefactor); dot((-3.7076164058457923,9.806497156043768),linewidth(4pt) + dotstyle); label("$R$", (-4.803168482667978,10.200992506496981), NE * labelscalefactor); dot((-9.544797136510528,-1.6279924300381627),linewidth(4pt) + dotstyle); label("$D$", (-10.687128672281088,-1.427149107245725), NE * labelscalefactor); dot((0.0890165433298842,-2.457507460144757),linewidth(4pt) + dotstyle); label("$E$", (-1.7041198014838537,-2.18327742836029), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We begin by noting that $R$ is the Miquel point of $MNBC$. Now note that there exists a spiral similarity with ratio $1$ taking $RMB \mapsto RNC$. In particular $(RNCK)$ has the same radii as $(RMBK)$, implying that $\widehat{MB} = \widehat{NC}$. Now let $D$ and $E$ denote the arc midpoints of $\widehat{MB}$ and $\widehat{NC}$ respectively. Clearly $D$ and $E$ lie on $PQ$, as $KQ$ bisects $\angle MKB$. Moreover it is easy to see that in the spiral similarity map $RMB \mapsto RNC$, we also have $D \mapsto E$ and that $Q$ is sent to the excenter of $\triangle KNC$. Then $QD = PE$. Now it suffices to show that $RD = RE$. However this is obvious as there exists a spiral similarity sending $MD \mapsto NE$ centered at $M$, due to the fact that $MN$ and $DE$ concur at $K$. Then noting by the spiral similarity sending $MN \mapsto BC$, we have $RM = RN$, we can conclude $RD=RE$ as desired. Thus we are done.

Let $R'$ be the center of the spiral similarity which takes $\overline{AM} \mapsto \overline{CN}$. By the length condition, this spiral similarity is just a rotation, so $R'A = R'C$. Moreover, $R'$ is also the center of the spiral similarity taking $\overline{AC} \mapsto \overline{MN}$, so $R'$ lies on $(ABC)$. This is enough to imply that $R' = R$ (the other intersection of the perpendicular bisector of $\overline{AC}$ with $(ABC)$ is the center of spiral similarity taking $\overline{AM} \mapsto \overline{NC}$). Now, since $R$ is the Miquel point of $AMNC$, we know that $KAMR$ and $KCNR$ are cyclic. So, if we let $M_1$ be the midpoint of arc $AM$ of $(KAMR)$ not containing $K$, and $M_2$ be the midpoint of arc $CN$ of $(KCNR)$ not containing $K$, the aforementioned spiral similarity takes $M_1 \mapsto M_2$, and hence $RM_1 = RM_2$. But by Incenter-Excenter lemma, we get $PM_1 = NQ_1$ because they're both equal to the radii of $(PAM)$ and $(CQN)$, which is enough to implty that $RP = RQ$.

We prove that $R$ is the Miquel point of $AMNC.$ Let $M', N'$ be points on $AB, BC,$ respectively, such that $RBMN$ is cyclic, and becaues of this, we have a spiral similarity sending $\triangle RMA ~ \triangle RNC.$ However, note that as $RA=RC$ because $R$ is the arc midpoint, this spiral similarity is actually a spiral congruency, so $RM=RN, AM=CN$ as well. This means that $R=(BNM) \cap (ABC),$ which is the definition of the Miquel point of $AMNC.$ \newline By Miquel properties, we have that $(KAMR)$ and $(KCNR)$ are cyclic quadrilaterals. Drawing these circles in, note that $\overline{PQ}$ is the angle bisector of $A$ due to incenter and excenter properties. Thus, the arc midpoints of $\overarc{MA}$ and $\overarc{CN}$ are also on $\overline{PQ}.$ Let these arc midpoints be $M_1, M_2,$ respectively. \newline Note that as $\overline{MN} \cap \overline{PQ},$ we have two spiral similarities sending $\triangle RMN ~ \triangle RM_1M_2,$ and $\triangle RMM_1 ~ RNM_2.$ This means that $RM_1=RM_2.$ As $\overline{PM_1M_2Q}$ is collinear, we now want to prove $PM_1=M_2Q.$ However, note that by incenter excenter lemma, we have that $M_1$ is the center of $(APM),$ and that $M_2$ is the center of $NQC.$ Note that the radii of these two circles are the same due to the spiral congruency sending $\triangle RMM_1 ~ \triangle RNM_2.$ Thus, $$PM_1=M_1M=NM_2=M_2Q,$$so $MP=NQ,$ and we're done.

Construct arc midpoints $J$ and $L$ opposite $K$ in $(AMK)$ and $(CNK)$, noting that $KPQJL$ collinear. This induces $\triangle AJM \cong \triangle CLN$. Since $R$ is the center of spiral congruence mapping \[\triangle RAM \mapsto \triangle RCN \implies RAJM \mapsto RCLN \implies RJ = RL.\] Moreover, incenter-excenter lemma tells us $JP = LQ$, which finishes. $\blacksquare$

$AR=CR,\angle RAB=\angle RCB,AM=CN$ give $\triangle RAM\cong\triangle RCN$ so $R$ is the miquel point of $ACNM.$ Now let the bisector of $\angle NKC$ meet $(KAM,KCN)$ at $D,F$ respectively. From $\measuredangle MDA=\measuredangle MKA=\measuredangle NKC=\measuredangle NFC$ and $AM=CN$ we get $\triangle MDA\cong\triangle NKC$ similarly oriented since both triangles are isosceles. Thus the rotation at $R$ taking $A$ to $C$ takes $D$ to $F$ thus $RD=RF.$ Since $PD=AD=CF=QF$ and $P,Q$ both lie outside segment $DF$ we have $RP=RQ.$

oml im actually braindead (didnt solve this is just a review) why didn't i draw $\overline{KPQ}\cap (KAM),(KCN)$? i'm actually so silly i decided to ignore this despite knowing fact 5 would probably do something. okay at least i thought of it so this is just me being really stupid anyway define $U,V$ as above. $\triangle AUM\sim \triangle CVN$ $RU=RV$ $PU=QV$ done

Consider the Miquel point of quadrilateral $AMNC$, $R'$. Because of the length condition, the spiral similarity is just a rotation, so $\triangle R'AM \cong R'NC$ implies $R'A = R'C$ and $R' \in \Gamma$ so $R' = R$. Then let $M_1$ and $M_2$ be the arc midpoints of $MA$ and $NC$. Note that $K - P - M_1 - M_2 - Q$, the spiral sim at $R$ sends $M_1 \to M_2$, and that $M_1P = M_2Q$. From this we have that $R$ lies on the perpendicular bisector of $M_1M_2$ and thus lies on the perp. bisector of $PQ$, as desired.

Note that since $CN=AM$, $R$ is the center of the rotation taking $NC$ to $MA$. The key idea of this problem is to "squish" the incenter and excenter together by the same amount along line $PQ$ to points $P'$ and $Q'$ and show that $RP'=RQ'$. We pick $P'$ to be the arc midpoint of $AM$ on $(RMAK)$, and $Q'$ to be arc midpoint of $CN$ on $(RNCK)$. First, we show that $PP'=QQ'$. Note that the rotation at $R$ sending $NC$ to $MA$ also sends $(RNCK)$ to $(RMAK)$, which means that $P'M=P'A=Q'N=Q'C$. By Fact 5, this means that $PP'=QQ'$. Finally, it suffices to show that $RP'=RQ'$. However, this simply follows by said rotation at $R$ sending $(RNCK)$ to $(RMAK)$, it also sends the arc midpoint to the corresponding arc midpoint.

Let $R'$ be the center of spiral similarity taking $\overline{CN}$ to $\overline{AM}$ So, $\triangle R'NC \cong \triangle R'MA \implies R'A=R'C$ Also, $R' \in (ABC)$ By configuration, we can see that, $R' \equiv R$ $\implies RM=RN$ and $(RMAK), (RNCK)$ Let $X=KP \cap (KXM)$ and $Y=KQ \cap (KCN)$ Thus, under the spiral similarity taking $\overline{CN}$ to $\overline{AM}$ centered at $R$, it is easy to see that $Y$ goes to $X$ $\implies RX=RY$ and $YQ=YN=XM=XP$ $\implies \triangle RYQ \cong \triangle RMP \implies RP=RQ$

I like my triangle $A$-indexed and hence I reform the labels a bit. Quote: Let $ABC$ be a triangle with $AC > AB$ and circumcircle $\Gamma$. Let $M$, $N$ be points on $AC$ and $AB$ with $BN = CM$. Let $K = \overline{MN} \cap \overline{BC}$. Denote by $P$ the incenter of $\triangle KCM$ and $Q$ the $K$-excenter of $\triangle BNK$. If $R$ is the midpoint of arc $BAC$ of $\Gamma$, prove that $RP = RQ$. See that $R$ is the Miquel point of $BCMN$. Let $L_1$ and $L_2$ be the minor arc midpoints in $\overarc{MC}$ and $\overarc{BN}$ in $(KCMR)$ and $(KBNR)$. See that since $\triangle RMC \overset{+}{\simeq} \triangle RNB$ and hence $\triangle RML_1 \overset{+}{\simeq} \triangle RNL_2$ and hence $RL_1=RL_2$. So we just need to prove $PL_1=QL_2$ but by Incenter-Excenter Lemma, this means we need to prove that $ML_1=NL_2$ which we just did.