A) let $B'$ the antipoode of $B$ in the circle $(O)$, $R$ is the centre of spiral similarity that takes $A$ , $B'$, $C$, $B$ to $K,S,P,B$ resp. since $RB'$ pass through the midpoint of $AC$ so $RS$ pass through the midpoint of $KP$

And $B)$:It's easy to prove that $CE \perp AB$ . (proof: $C$ lies on the radical center of $(AN),(BM)$ and midline of $AMNB$ , $AB$ are parallel). $MQ$ cuts $CE$ at $H$ so $H$ be the orthocenter of $\triangle ABC$. $RK$ cuts $CE$ at $Z$. $KP \parallel CE$ so $\angle RZC=\angle RKP=\angle CBR$ so $Z$ lies on $(ABC)$. by $A)$ : $SR$ passes through the midpoint of $KP$ so $E$ be the midpoint of $ZH$ thus $E$ lies on $AB$ so $E$ lies on $\odot (N,NB)$ .assume that $ME$ cuts $\odot (N,NB)$ at $X$ then we have : $\angle XEC=\angle MEC=\angle MCE=\angle OBC$ so $X$ lies on $BO$ . Thus $ME$ always passes through $X$ (the intersection of $\odot N$ and $BO$) . done.

Solution ($a $): Observe that the circle with diameter $BP $ also passes through $R $. Also, $B $, $S $ and $Q $ are collinear. Let $PK\cap SR = X $. Join $QA $ and $QC $. It would be enough to show that $PR\cdot \sin\angle PRS = KR\cdot \sin\angle KRS $. We have, $\frac {PR}{KR} = \frac {\sin \angle PKR}{\sin \angle RPK} = \frac {\sin \angle RQC}{\sin \angle AQR} = \frac {\sin \angle ACQ}{\sin \angle CAQ} = \frac {\sin \angle ABQ}{\sin \angle CBQ} = \frac {\sin \angle KRS}{\sin \angle PRS} $. This completes the proof.

Here is my solution for this problem Solution a) Let $H$ be orthocenter of $\triangle$ $ABC$; $V$ be midpoint of $KP$ Since: ($RQ$; $RB$) $\equiv$ ($RM$; $RB$) $\equiv$ $\dfrac{\pi}{2}$ (mod $\pi$) then: $Q$ is reflection of $B$ through $O$ So it's well - known that: $H$ $\in$ $MQ$ and $M$ is midpoint of $HQ$ We have: ($PK$; $PR$) $\equiv$ ($BK$; $BR$) $\equiv$ ($QA$; $QR$) $\equiv$ ($HC$; $HQ$) $\equiv$ $-$ ($HQ$; $HC$) (mod $\pi$) and ($KR$; $KP$) $\equiv$ ($BR$; $BP$) $\equiv$ ($QR$; $QC$) $\equiv$ ($QH$; $QC$) $\equiv$ $-$ ($QC$; $QH$) (mod $\pi$) Hence: $\triangle$ $KPR$ $\stackrel{-}{\sim}$ $\triangle$ $QHC$ But: $I$, $M$ are midpoint of $KP$, $QH$, respectively then: $\triangle$ $VPR$ $\stackrel{-}{\sim}$ $\triangle$ $MHC$ So: ($RV$; $RS$) $\equiv$ ($RV$; $RP$) + ($RP$; $RS$) $\equiv$ ($CH$; $CA$) + ($BC$; $BQ$) $\equiv$ ($CH$; $CA$) + ($BH$; $BA$) $\equiv$ ($BA$; $BH$) + ($BH$; $BA$) $\equiv$ 0 (mod $\pi$) or $S$, $V$, $R$ are collinear Therefore: $RS$ passes through midpoint $V$ of $KP$ b) We define again point $E$ $\equiv$ $HC$ $\cap$ $AB$ Let $D$ $\equiv$ $AH$ $\cap$ $BC$; $F$ $\equiv$ $BH$ $\cap$ $CA$ We have: $P_{H / \bigodot(AN)}$ = $\overline{HA}$ . $\overline{HD}$ = $\overline{HB}$ . $\overline{HF}$ = $P_{H / \bigodot(BM)}$ and $P_{C / \bigodot(AN)}$ = $\overline{CD}$ . $\overline{CN}$ = $\overline{CF}$ . $\overline{CM}$ = $P_{C / \bigodot(BM)}$ Then: $HC$ is radical axises of $\bigodot(AN)$ and $\bigodot(BM)$ But: ($RS$; $RE$) $\equiv$ ($RS$; $RP$) + ($RP$; $RE$) $\equiv$ ($CH$; $CA$) + ($BC$; $BQ$) $\equiv$ ($CH$; $CA$) + ($BH$; $BA$) $\equiv$ ($BA$; $BH$) + ($BH$; $BA$) $\equiv$ 0 (mod $\pi$) so: $E$, $S$, $R$ are collinear Let $U$ be orthogonal projection of $C$ on $BQ$ then $U$ is fixed point We have: ($EU$; $EM$) $\equiv$ ($EU$; $EC$) + ($EC$; $EM$) $\equiv$ ($BQ$; $BC$) + ($CA$; $CE$) $\equiv$ ($BA$; $BH$) + ($BH$; $BA$) $\equiv$ 0 (mod $\pi$) So: $E$, $U$, $M$ are collinear or $EM$ passes through fixed point $U$

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Vietnam TST 2017 P4l wrote: Triangle $ABC$ is inscribed in circle $(O)$. $A$ varies on $(O)$ such that $AB>BC$. $M$ is the midpoint of $AC$. The circle with diameter $BM$ intersects $(O)$ at $R$. $RM$ intersects $(O)$ at $Q$ and intersects $BC$ at $P$. The circle with diameter $BP$ intersects $AB, BO$ at $K,S$ in this order. a. Prove that $SR$ passes through the midpoint of $KP$. b. Let $N$ be the midpoint of $BC$. The radical axis of circles with diameters $AN, BM$ intersects $SR$ at $E$. Prove that $ME$ always passes through a fixed point. Solution: $\angle BKP=\angle BSP=\angle BRP=90^{\circ}$ $\implies$ $R$ $\in$ $\odot (BKSP)$. Also, $Q$ coincides with $B'$, the $B-$antipode. Also, $RM$ passes through orthocenter $H$ WRT $\Delta ABC$. Let $E,F$ be the foot from $B,C$. Let $BH$ $\cap$ $\odot (BRH)$ $=$ $G$. Since, $BS,BG$ are isogonal WRT $\angle ABC$ $\implies$ $SG||KP$. Already $KP||CF$ $\implies $ By converse of Reim's Theorem $F$ $\in$ $SR$. $$-1=(A,C; BH ~ \cap ~\odot (ABC), R) \overset{B}{=} ( K,P ; G ,R) \overset{S}{=} ( K,P ; \infty_{KP}, SR ~ \cap ~KP)$$Hence, $SR$ bisects $KP$. Let $D$ be the foot from $A$. Since, $$CD \cdot CN=CE \cdot CM ~ \& ~ AH \cdot HD= BH \cdot HE$$Hence, $CH$ is the radical axis WRT $\odot (AN)$ and $\odot (BM)$. Let $\odot (BFC)$ $\cap$ $BB'$ $=$ $X'$ $$\angle X'FC=\angle SBP=\angle SKP=\angle MFC$$Hence, $X' \in MF$. Note, $CH \cap SR=F$ and $X'$ is fixed $\qquad \blacksquare$