My solution: Let $ D=AH \cap BC, E=BH \cap CA $ . It's well-known that $ CS $ is the isogonal conjugate of $ CM $ WRT $ \angle ACB $ , so combine with $ \triangle CAD \sim \triangle CBE \Longrightarrow \triangle CAD \cup S_a \cup M_a \sim \triangle CBE \cup M_b \cup S_b $ $ (\star) $ , hence we get $ \angle DM_aC=\angle ES_bC \Longrightarrow M_a, M_b, S_a, S_b $ are concyclic . From Steiner theorem and $ (\star) \Longrightarrow \frac{AC^2}{AS_a \cdot AM_a}=\frac{DC^2}{DS_a \cdot DM_a} =\frac{EC^2}{ES_b \cdot EM_b}=\frac{BC^2}{BS_b \cdot BM_b} $ , so $ \odot (ABDE) $ is coaxial with $ \odot (S_aS_bM_aM_b) $ and the degenerate circle $ C $ , hence the center of $ \odot (S_aS_bM_aM_b) $ lie on the line connecting $ C $ and the center $ M $ of $ \odot (ABDE) $ , so we get $ M_aM_b $ is the diameter of $ \odot (S_aS_bM_aM_b) $ and $ M_aS_b \perp HM_b, M_bS_a \perp HM_a $ . Q.E.D

TelvCohl wrote: My solution: Let $ D=AH \cap BC, E=BH \cap CA $ . It's well-known that $ CS $ is the isogonal conjugate of $ CM $ WRT $ \angle ACB $ , so combine with $ \triangle CAD \sim \triangle CBE \Longrightarrow \triangle CAD \cup S_a \cup M_a \sim \triangle CBE \cup M_b \cup S_b $ $ (\star) $ , hence we get $ \angle DM_aC=\angle ES_bC \Longrightarrow M_a, M_b, S_a, S_b $ are concyclic . From Steiner theorem and $ (\star) \Longrightarrow \frac{AC^2}{AS_a \cdot AM_a}=\frac{DC^2}{DS_a \cdot DM_a} =\frac{EC^2}{ES_b \cdot EM_b}=\frac{BC^2}{BS_b \cdot BM_b} $ , so $ \odot (ABDE) $ is coaxial with $ \odot (S_aS_bM_aM_b) $ and the degenerate circle $ C $ , hence the center of $ \odot (S_aS_bM_aM_b) $ lie on the line connecting $ C $ and the center $ M $ of $ \odot (ABDE) $ , so we get $ M_aM_b $ is the diameter of $ \odot (S_aS_bM_aM_b) $ and $ M_aS_b \perp HM_b, M_bS_a \perp HM_a $ . Q.E.D I don't understand ,why circles are coaxial?

izat wrote: I don't understand ,why circles are coaxial? $ \frac{AC^2}{AS_a \cdot AM_a}=\frac{DC^2}{DS_a \cdot DM_a} =\frac{EC^2}{ES_b \cdot EM_b}=\frac{BC^2}{BS_b \cdot BM_b} $ means the ratio of the power of $ A, B, D, E $ WRT $ \odot C $ and $ \odot (S_aS_bM_aM_b) $ are the same , so $ A, B, D, E $ lie on a circle coaxial with the degenerate circle $ \odot C $ and $ \odot (S_aS_bM_aM_b) $ .

Let $D,E$ be the feet of the altitudes on $CB,CA.$ $Z$ is the antipode of $C$ on $w$ and $CM$ cuts $AZ$ at $L.$ Since $CDHE \cup S_a \sim CAZB \sim L$ $\Longrightarrow$ $\tfrac{HS_a}{S_aD}=\tfrac{ZL}{LA}.$ But from parallelogram $HBZA$ with center $M,$ we obtain $\tfrac{ZL}{LA}=\tfrac{HM_b}{M_bB}$ $\Longrightarrow$ $\tfrac{HS_a}{S_aD}=\tfrac{HM_b}{M_bB}$ $\Longrightarrow$ $M_bS_a \parallel BC$ $\Longrightarrow$ $M_bS_a \perp HM_a$ and similarly $M_aS_b \perp HM_b.$

SOLUTION (WITH amar_04 and me) Let E,F be the projection of $B,C$ onto $AC$,$AB$ respectively. Let $EF\cap BC=K$. As BEFC is cyclic quadrilateral, by brocard theorem We get that $KH\perp AM$ at $I$, LET $BE\cap AM=M_b$,and $AH\cap BC=D$. Now $$-1=(D,K;B,C)\overset{H}{=}(A,I;M_b,M_a)$$Now lets draw, $M_bS_a\perp CF$ And $M_aS_b\perp BE$, Then it is enough to show that $AS_a$ is symmedian.Now $A,S_b,S_a$ must be collinear( if $S_aS_b\cap AM=P$, THEN (P,I;$M_b$,$M_a$) will be harmonic, contradiction). Now it is easy to see that $\angle M_bAE=\angle M_aHI$ and $\angle FAM_a=\angle M_bM_aS_b=\angle M_bHI$, SO A$S_a$ is symmedian