In a $\Delta ABC$ prove: $\angle AST=\angle OSB$ if $T$ is the midpoint of $AH$ and $S=BC \cap t$ ($t$ = tangent line in $A$ to a circumcircle) 
Let $ABC$ a triangle and $H$ and $O$ its orthocenter and circumcenter. Denote by $T$ the midpoint of the segment $AH$ and by $S$ the intersection of the tangent in $A$ to the circle $ABC$ and the line $BC$. Prove that $\angle AST=\angle OSB$

Let $T$ be a midpoint of the segment $BC$ and $D$ an orthogonal projection of $A$ on $BC$.


*

*Since $TEOA$ is parallogram we see that $T$ is an orthocenter in triangle $ASE$, so $$\angle SEA = \angle STD $$

*Since $OE\bot BC$ we see that $SEOA$ is cyclic, so $$\angle SEA = \angle SOA $$
Putting this together we get $\angle SOA = \angle STD\implies \angle OSA = \angle TSD $ and thus a conclusion.
I'm interested in projective or/and inversion solution. 
 A: We have a simple proof, but accepting the challenge we may reveal more structure.
So let us show:

Let $\Delta ABC$ be a triangle. Its medians are as a matter of notation $AA'$, $BB'$, $CC'$, its heights $AA_1$, $BB_1$, $CC_1$. Let $O$ be the center of the circumcircle $(ABC)$, let $H$ be the orthocenter. We also distinguish the mid points $A^*$, $B^*$, $C^*$ of the segments $AH$, $BH$, $CH$ respectively. Then the nine points $A',B',C';A_1,B_1,C_1;A^*,B^*,C^*$ are on a circle $(9)$ centered in $9$, the mid point of $OH$, since it projects on the sides on the mid points of its chords $A'A_1$, $B'B_1$, $C'C_1$.
Then $OA'$, $B^*C'$, $C^*B'$, $AA^*$, $A^*H$ are equal segments supported by parallel lines. In particular, $AA^*A'O$ is a parallelogram.

We consider now the inversion $\mathcal I$ with center $A$ that invariates the circle $(9)$ as a set, the power of this inversion being
  $$
k^2=AA^*\cdot AA_1=AB'\cdot AB_1=AC'\cdot AC_1
\ .
$$

Let $U=\mathcal I A'$. Then $A^*U\perp AA'$. 
Let $V$ be $AS\cap AA'$. Then $V=\mathcal I S$.
The points $S,A^*,U$ are colinear. So $A^*$ is the orthocenter of $\Delta ASA'$, and its heights are $AA_1$, $SU$, $A'V$.

The line $SO$ is perpendicular on $A_1V$, and we have finally the equality of angles:
  $$
\begin{aligned}
\widehat{ASA^*} 
&=
\widehat{AA'A^*} = \widehat{AA'V} \\
&=
\widehat{AA_1V} \\
&=
\widehat{OSB} \ .
\end{aligned}
$$

Proof:


*

*$OA'B^*C'$ is a parallelogram, because $A'B^*\| CH =CC_1\|OC'$, and similarly $C'B^*\| AH =AA_1\|OA'$. This gives $OA'=C'B^*=\frac 12AH=AA^*$. It follows $AA^*A'O$ parallelogram. We will use only $A^*A'\|A'O$, both lines being thus perpendicular on the tangent $AS$.

*The inversion $\mathcal I$ transforms $A^*\leftrightarrow A_1$, and 
$A'\leftrightarrow U$. (The last by the definition of $U$.) So $A^*A_1A'U$ cyclic. The angles in $U$, $A_1$ add to $\pi$, so we have a right angle in $U$, i.e. $A^*U\perp AA'$. By construction, we also have a right angle in $V$ in $AVA^*U$, which is transformed by $\mathcal I$ in the line $\infty (\mathcal IV)A_1A'$. So $\mathcal IV$ lies on both lines $AS$, and $A_1A'=BC$, so it is their intersection, $S$. The points $S,A^*,U$ are colinear, because their transformed points $V$, $A_1$, $A'$ lie on a circle passing through $A$. (Because we have right angles in $V$, $A_1$ in $VA_1A'A$.)  

*Let us consider now the inversion $J$ centered in $S$ with power $SV\cdot SA =SA_1\cdot SA'$. It invariates $SO$. The line $VA_1$ is transformed in the circle $SAA'$ which passes through $O$, because $SAOA'$ cyclic, because of the two right angles in $A,A'$. Moreover, and for this reason, $SO$ is a diameter in the circle $(SAOA')$. This means that $SO=\mathcal J(SO)$ intersects orthogonally in $O$ the circle $(SAOA')=\mathcal J(VA_1)$. The inversion preserves angles, so $SO\perp VA_1$.

*From $SO\perp VA_1$ and $SA_1\perp AA_1$ we obtain equal angles between the two pairs of perpendicular lines, this explains the last step in
$$
\begin{aligned}
\widehat{ASA^*} 
&=
\widehat{AA'A^*} = \widehat{AA'V} &&UVSA'\text{ cyclic}\\
&=
\widehat{AA_1V} &&AA'A_1V\text{ cyclic}\\
&=
\widehat{OSB} &&\text{ angles between pairs of $\perp$ lines.}
\end{aligned}
$$
$\square$

Note: A proof by inversion, or a proof inside projective geometry is difficult because of the fact that there is "no further distinguished point" on $OS$. The idea of proof above was to use the "distinguished perpendicular" $VA_1$ on it. The proof does not need all points in the picture, but i inserted them all, it is a good feeling to see ten points on a circle. But also because there are some further properties in the given picture, for instance...
Bonus:
$B_1C_1$ is perpendicular on $A'A*V\|OA$, and is thus parallel to the tangent $AS$. (This is because of the equality of chords $A^*B_1=A^*C_1$ in the circle $(9)$, both being half of the hypothenuse $AH$.) Further, let $B_1^9$ be the diametraly opposed point in $(9)$ for $B_1$. Then $B_1^9C_1\|A'A^*V\|OA$, these are perpendicular lines on the direction given by $AS\|B_1C_1$. The point $B_1^9$ was constructed so that $\Delta B_1C_1B_1^9$ has the right right angle, this would be a cheapo, but thematically we also have: $B_1^8C_1$, $AO$, and $A'B^*$ are concurent. The point $(9)$ was introduced, so that we can finally mention: $B9$, $AA_1$ and $A'B^*$ are concurent. 
