$$\triangle ABC$$ $$AA_1, BB_1, CC_1 - h$$ $$AA_1 \cap BB_1 \cap CC_1 = H $$ $k(O; r)-$ circumcircle of $\triangle ABC$
$k_1(0_1; r_1) -$ circumcircle of $\triangle AHB$
Show that $k \cong k_1$.
$A, B \in k$, thus $OA=OB=r$, thus $O \in S_c$
$A, B \in k_1$, thus $O_1A=O_1B=r_1$, thus $O_1 \in S_c$, thus $OO_1 \equiv S_C$
$OO_1 \bot AB, OO_1 \cap AB = P : AP=PB$.
I don't know how from here to show that $AO = AO_1$. Thank you in advance!
Easier: reflect $H$ in $AB$ to a point $H'$. We will show that $H'$ lies on the circumcircle of $ABC$, hence the circumradius of $ABC=ABH'$ and $ABH$ are equal.
We have $\angle HAB=90^\circ-\angle ABC$, $\angle HBA=90^\circ-\angle CAB$. Hence $$ \angle AH'B= \angle AHB=180^\circ-\angle HAB-\angle HBA =\angle ABC + \angle CAB = 180^\circ - \angle ACB $$ so $AH'BC$ concyclic, i.e., $H'$ lies on the circumcircle of $ABC$.