As $M$ is midpoint of arc $AD$ then for every point $X$, that lies on the arc $AD$ not containing $M$ it is truem that $XM$ is angle bisector of $AXD$ Let $QC$ intersect $ MC$ at $G$. $\angle APB=180-\angle BAP -\angle ABP$. $\angle NMC= \frac{\angle BAC}{2},\angle PCM=\frac{\angle ABD}{2}$, $\angle PGM=90- \angle NMC= 90-\frac{\angle BAC}{2}, \angle CPG=\angle PGM-\angle PCG= 90-\frac{\angle BAC}{2}-\frac{\angle ABD}{2}= \frac{CPD}{2}$ so $GP$ is angle bisector of $CPD$ and $G$ - is incenter of $CPD$ $QP$ is angle bisector of $APB$ and $Q$ lies on $BM$ - angle bisector of $ABP$, so $Q$ is incenter of $APB$ Let $F$ is point of intersection of $AC$ and $QR$. $QF= \frac{AQ \sin \angle NAC}{\sin \angle QFA}=\frac{BQ \sin \angle ABM }{\sin \angle QFA}=\frac{RC \sin \angle ACM }{\sin \angle CFR}=FR$

Let $X=QR\cap AC$, by menelao in $\triangle QMR$ $$\frac{MC}{CR}\cdot \frac{RX}{XQ}\cdot \frac{QY}{YM}=1$$ So is sufficient to prove that $$\frac{MC}{CR}=\frac{YM}{QY}$$ As $\angle BMN=\angle NMC, \angle MBD=\angle ACM$ by exterior angle it´s easy to see that $PQ$ is the bisector of $\angle BPA$ then $$\frac{BQ}{QY}=\frac{PB}{PY}$$ Note that $\frac{BP}{PY}=\frac{\sin\angle BYP}{\sin \angle PBY}=\frac{\sin\angle CYM}{\sin\angle YCM}=\frac{MC}{MY}$, so $\frac{BQ}{QY}=\frac{MC}{MY}$ or $\frac{MC}{CR}=\frac{YM}{QY}$, as desired.

RagvaloD wrote: As $M$ is midpoint of arc $AD$ then for every point $X$, that lies on the arc $AD$ not containing $M$ it is truem that $XM$ is angle bisector of $AXD$ Let $QC$ intersect $ MC$ at $G$. $\angle APB=180-\angle BAP -\angle ABP$. $\angle NMC= \frac{\angle BAC}{2},\angle PCM=\frac{\angle ABD}{2}$, $\angle PGM=90- \angle NMC= 90-\frac{\angle BAC}{2}, \angle CPG=\angle PGM-\angle PCG= 90-\frac{\angle BAC}{2}-\frac{\angle ABD}{2}= \frac{CPD}{2}$ so $GP$ is angle bisector of $CPD$ and $G$ - is incenter of $CPD$ $QP$ is angle bisector of $APB$ and $Q$ lies on $BM$ - angle bisector of $ABP$, so $Q$ is incenter of $APB$ Let $F$ is point of intersection of $AC$ and $QR$. $QF= \frac{AQ \sin \angle NAC}{\sin \angle QFA}=\frac{BQ \sin \angle ABM }{\sin \angle QFA}=\frac{RC \sin \angle ACM }{\sin \angle CFR}=FR$ Very nice! Just a small typo: $G\in PQ\cap MC$. To avoid use of trigo calculations, I ended your proof as follows: $\triangle APB\sim\triangle DPC, Q$ and $R$ are their respective incenters, thus $\frac{QP}{GP}=\frac{QB}{GC}=\frac{CR}{GC}$ and, from Menelaos in $\triangle CQG$ with transversal $Q-F-R$ we get $FR=FG$. Best regards, sunken rock

Evidently $Q$ is the incenter of $PAB$. Hence we may delete some points and reduce the problem to the following: new problem wrote: Consider a triangle $MBC$, and let $S$ and $N$ denote the midpoints of the major and minor arcs $BC$. Two points $Q$ and $R$ are chosen on segments $MB$ and $MC$ such that $QB = RC$. If $T$ and $K$ are the respective midpoints of $QR$ and $BC$, prove that $TC$ and $QN$ meet on $(BMC)$.

By angle chasing we get $Q$ is the incenter of $PAB$. Then $\angle BAQ= \angle BAC/2=\angle BAN$. So $A, Q, N$ collinear. In a similar way we let $Q'=MC\cap PQ$. $Q'$ is the incenter of $PDC$. So $D, Q', N$ collinear. As $\angle ANM = \angle MND$ and $NM \perp QQ'$ we get $NM$ is angle bisector of $QQ'$. By angle chasing we get $\triangle QBP$ is similar to $\triangle Q'CP$. Now let $R'$ on $AC$ such that $RR' \parallel QQ'$. Then $\triangle RCR'$ is also similar, as $QB=RC$ then $RR'=QP$ and since these are parallel they form a parallelogram. So $QT$ and $PR'$ intersect at their midpoints. In particular $T=PR'\cap QR$ then T lies on AC and is the midpoint of QR. $\square$