Suppose $AC>AB.$ Let $D$ be the midpoint of $AB$ and $H$ be the orthocenter of $ABC.$ Draw line $AO$ and mark its intersection with the circle at $A'.$ Then $\angle HBC=\angle ABO=\angle BAO=\angle BCA',$ so $HB$ and $CA'$ are parallel. By symmetry, $HC$ and $A'B$ are parallel. Therefore $HCA'B$ is a parallelogram with center $D.$ Then draw our second line $AD,$ which goes through $H.$ Now draw diameter $DO$ and this line intersects the triangle on $AC$ at point $E.$ Then draw $BE$ and this line intersects the circle at $F,$ which is $A$ reflected about $DO.$ Draw line $FO$ to intersect the circle at $G$. Then $AG$ has an altitude of $ABC.$ So the orthocenter of $ABC$ is the intersection of $AG$ and $AD.$