Given an acute triangle in a circle, prove the concyclicity of some points Given an acute triangle ABC with AB<AC<BC written on a circle c(O, R). The extension of the height AD intersects the circle c(O, R) at E. The circle c1(D, DA) intersects the side AC at the point T, the line AB at the point S, the circle c(O, R) at H and the line OA at Z. Prove that the points S, B, T, C are concyclic on a circle c2 and that the points O, D, E, Z, H and the center of c2 are concyclic.

I attempted to do the question by saying that: If AD intersects the circle c1 at P, then AP is the diameter of c1, hence ASP=ATP=90.
Since BSP=BDP=90 and PDC=PTC=90, then the points B, D, P, S are concyclic, as are the points D, T, C, P. This is where I got stuck. Could you please explain to me how to finish off this question?
 A: 
Observe that, $\angle BST=\angle AST=\angle APT=90-\angle PAT=90-\angle DAC=\angle C=\angle TCB$.    $\Rightarrow$   Points $B$, $S$, $C$ and $T$ are concyclic.
The perpendicular bisectors of $BC$ and $ST$ meet at $R$;Hence, it is the centre of circle $C_2$. The perpendicular bisectors of $BC$ and $ST$ will pass through point $O$ and $D$ respectively. Let $AZ$ intersect $ST$ at $K$.
Observe that, $\angle SAK+\angle ASK=\angle BAO+\angle AST=90-\angle C+\angle C=90$
$\Rightarrow \angle AKS=90^{\circ}$
$\Rightarrow  AK\parallel DR$
Since $DR\parallel AO$ & $AD\parallel OR$, $AORD$ is a parallelogram.
$\Rightarrow \angle DAO=\angle DRO$
Now, notice that, $\angle DZO=\angle DAO$[ Since $DZ=DA$]$=\angle DRO$; Hence, Points $D$, $O$, $Z$ and $R$ are concyclic.
Also, observe that, $\angle DEO=\angle DAO$[ Since $OE=OA$]$=\angle DRO$.
$\Rightarrow$  Points $D$, $E$, $R$ and $O$ are concyclic.
Since $DH=AD$ and $AO=OH$, $\angle DHO=\angle DAO$
$\Rightarrow   \angle DHO=\angle DAO=\angle DRO$.
$\Rightarrow$  Points $D$, $O$, $H$ and $R$ are concyclic.
Therefore, points $E$, $D$, $O$, $H$, $Z$ and $R$, which is the centre of circle $C_2$, are all concyclic.
