Let $D$ be the antipode of $C$ in $\omega$. Note that $PQ \parallel AB$ so $CH \perp PQ \Longrightarrow P, Q, R, S$ are concyclic. Since $M$ is the midpoint of $DH$ and $\angle CQD=\angle CPD=90^{\circ}$ we obtain that $M$ lies on the perpendicular bisectors of $SP$ and $QR$. It follows that $M$ is the center of the circle passing through $P, Q, R, S$.

Oops what is antipode ?

Lemma 1: $PQRS$ is cyclic. Proof: We use power of $C$ w.r.t. $PQRS$. We denote $ <PCA=\phi$. Namely, $CR=CH \sin(\alpha +\phi)$, and $CQ=2R\sin(\beta+ \phi)$. So, $CS \cdot CQ = 2R \cdot CH \sin(\alpha +\phi)\sin(\beta+\phi)$, clearly symetric. Let $D$ and $E$ be the foots of altitudes from $A$ and $B$. Lemma 2. $M$ is the center of the circle. Proof: Bisectors of $AB$ and $PQ$ coincide, and bisectors of $SR$ and $ED$(because $CSRHED$ is cyclic), and so $M$ lies on both bisectors, so it's the center of the circle.

kk108 wrote: Oops what is antipode ? Diametrically opposite point

Let $CH\cap PQ=K$ İn order to get $PQRS$ is cyclic we will use PoP Since $AP=BQ$ $\implies$ $AB$ is parallel to $PQ$.Little angle-chasing gives $SHPK$ and $HRQK$ are cyclic $\implies$ $CS\cdot CP=CH\cdot CK=CR\cdot CQ$ $\implies $ $PQRS$ is cyclic. Now triangles $AMP$ and $BMQ$ are.congruent $\implies$ $MP=MQ$ Now using this famous lemma done!!! İn triangle $ABC$ $HM$ $CC'$ intersect on $(\odot ABC)$ and $M$ is midpoint of $C'H$ where $H,M,C'$ are orthocenter,midpoint of $AB$ and antipode of $C$ respectively

tarzanjunior wrote: kk108 wrote: Oops what is antipode ? Diametrically opposite point Thank you !

Who knows proposer of the problem?

Juraev wrote: Who knows proposer of the problem? M.Kungozhin

My solution : Let $\angle ACP=x$ we have that $CS=CH \cos{90-a-x},CR=CH\cos{90-b-x}$ so we have that $\frac{ CR}{CS}=\frac{\sin{b+x}}{\sin{a+x}}$. We have that $\angle QPC=a+x$ and $\angle PQC=b+x$ from sine law we have $\frac{CR}{CS}=\frac{CP}{CQ}$. Therefore $PQRS$ is cyclic. Since $CHRS$ is cyclic we have that $XN\perp RS$ where $X,N$ are midpoints of $CH,RS$. But $XM\parallel CO$ where $O$ is circumcircle and $CO\perp RS$ because $\angle (RS,BC)=a$ so $XM\perp RS$ which $M$ is bisector of $RS$. Because $\triangle MPA=\triangle MQB$ $M$ is in bisector of $PQ$ which implies that $M$ Is the center.

Let $D$ and $E$ be the foot of the altitudes from $A$ and $B$, respectively. Let the circumcircle of $CDE$ be $t$. It is easy to see that $CDRHSE$ is cyclic because $R$ and $S$ are foot of the altitudes from H. Because $\angle CBQ=\angle ACP$, $CDE$ is similar to $CAB$ and $R$ and $S$ belong to $t$, we can see that $CRS$ is similar to $CPQ$, so $CR*CQ=CS*CP$, so first part is done. For the second part, see that $BQ=PA$ and $DR=SE$ because that pairs belong to $\omega$ and $t$, respectively, and the angle condition. Finally, because $BQPA$ and $DRSE$ are isosceles trapezoids, the perpendicular bisector of $QP$ and $SR$ are the some of $BA$ and $ED$, respectively. But the perpendicular bisector of $DE$ passes through $M$ (because $DME$ is isosceles), so $M$ is the center of the circumcircle of $PSRQ$.

I checked real zautykov problem list.There is written that ABC IS NOT ISOSCELES TRIANGLE

There is easy solution by complex numbers. Let $X$ be the feet of altitude from $M$ to $CQ$. For solving problem we need to prove that $X$ is the midpoint of $QR$. $4x=2r+2q$ $2c+2q+2m-2cq\overline{m}=c+q+h-cq\overline{h}+2q$ $q+2cq\overline{m}=cq\overline{h}$ $1+2c\overline{m}=c\overline{h}$ $1+c(\frac{1}{a}+\frac{1}{b})=c(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ and we are done.