Let $H$ and $O$ be the orthocentre and the centre of the circumcircle of $\triangle ABC$ respectively, and $E$ be the centre of $\omega$. Then, as you know $E$ be the centre of $HO$. Let $M$ be the midpoint of $BC$. Let the circle $\Omega$ intersect $BC$ for the second time at point $X$. Since $\angle CAB = 60^\circ$, then $\angle BHC = \angle BOC = 120^\circ$, so points $B, H, O$ and $C$ are on the same circle (fig. 23). Let $T$ be the point of intersection of the lines $UV$ and $BC$, then, according to the degree of the point, $TD \cdot TX = TU \cdot TV = TB \cdot TC = TH \cdot TO$. Thus, the points lie on the same circle, and since $\angle HDX = 90^\circ$, then $\angle HOX = 90^\circ$, whence $XO \perp HO$. And remembering that $OU = OV$, we get that $XO$ bisects chord $UV$ of the circle $\Omega$, from where $XO$ passes through the centre of the circle $\Omega$, and hence $XO$ passes through the point diametrically opposite to the point $X$ on the circle $\Omega$. However $\angle XDA = 90^\circ$, hence $Z$ is the point diametrically opposite to $X$ on the circle $\omega$. So $XO$ passes through $Z$. In the right $\triangle BOM$, $\angle BOM = \frac{1}{2}\angle BOC = \angle BAC = 60^\circ$, so

$AH = BO = AO$. Thus, the condition $\angle BAC = 60^\circ$ implies that $AH = AO$, and therefore $AE \perp HO$, with $AE$ being the bisector $\angle HAO$, and hence $AE$ is also the bisector of $\angle CAB$. The condition $\angle BAC = 60^\circ$ implies that the radius of the circle $\omega$ is $\frac{1}{2}AO = \frac{1}{2}AH = OM$. Since $\angle HOZ = 90^\circ$ the condition $AH = AO$ implies that $A$ is the midpoint of the segment $HZ$. Denote by $W$ the point that is symmetric to $O$ with respect to $M$. Then $W$ is also symmetric to $O$ with respect to $BC$, since $OM \perp BC$, so $BW = WC$ and $\angle BWC = 120^\circ$, whence $W$ is the midpoint of the smaller arc $BC$ of the circumcircle $\triangle ABC$, and therefore $W$ lies on $AE$. Moreover, since $OA = OW$ and $OE \perp AW$, then $E$ is the midpoint of the segment $AW$. In the trapezoid $ZOWH$, $A$ and $M$ are the midpoints of the bases $ZH$ and $OW$, so, by a well-known fact, the point of intersection of its diagonals $ZW \cap HO$ lies on $AM$, but $AM \cap HO = G$ is the centroid of $\triangle ABC$, so $G$ lies on $ZW$. Let us prove the collinearity of the points $Z, K$ and $W$, which implies that $G, K, Z$ and $W$ lie on the same line. To do this, let $K'$ be the projection of point $X$ onto $ZW$. Moreover, since $OA=OW$, and $OE \perp AW$, $E$ is the midpoint of the segment $AW$. In the trapezoid $ZOWH$, $A$ and $M$ are the midpoints of the bases $ZH$ and $OW$, so, by a well-known fact, the point of intersection of its diagonals $ZW \cap HO$ lies on $AM$, but $AM \cap HO=G$ is the centroid of $\triangle ABC$, so $G$ lies on $ZW$. Let us prove the collinearity of the points $Z, K$ and $W$, which implies that $G, K, Z$ and $W$ lie on the same line. To do this, let $K'$ be the projection of point $X$ onto $ZW$. Then, since, $\angle XK'Z = \angle XDZ = 90^\circ$, the points $X, Z, D$ to $K'$ are cyclic, and therefore $K'$ lies on the circle $\Omega$. However $K'$ lies on $ZW$, so to prove the collinearity of the points $Z, K$ and $W$, it is enough to show that $K'$ lies on the circle $\omega$, which implies that $K'$ is the second intersection of the circles $\Omega$ and $\omega$, i.e. $K' = K$, and therefore $K$ like $K'$ will lie on the line $ZW$. We have that, $\angle WK'X = \angle WMX = 90^\circ$, so the points $X, M, K'$ and $W$ lie on the same circle, and hence $\angle MK'X = \angle MWX = \angle MOX$ (this equality is obtained from the axial symmetry of points $O$ and $W$ with respect to line $BC$). However, the two lines $ZD$ and $OM$ are perpendicular to $BC$, so they are parallel, so $\angle MOX = \angle DZX$, i.e. $\angle MK'X = \angle DZX$. Since the points $X, Z, K'$ and $D$ are cyclic, then $\angle DK'X = 180^\circ - \angle DZX$. Therefore
$$\angle DK'M = \angle DK'X - \angle MK'X = 180^\circ - \angle DZX - \angle DZX = 180^\circ - 2\angle DZX.$$

$$\angle DK'M = 180^\circ - 2\angle DZX = 180^\circ - 2\angle DAE = 180^\circ - \angle DAO.$$
Let $N$ be the midpoint of $AH$, then, as we know, $N$ lies on $\omega$, and $\angle MDN$ is right, so $MN$ is the diameter of the circle $\omega$, so the points $M, E$ and $N$ are collinear. Moreover, since $AN = \frac{1}{2}AH = OM$ and $AH\parallel OM$, then $AOMN$ is a parallelogram, hence $MN\parallel AO$, and therefore $\angle DAO = \angle DNM$. Thus, as we proved above, $\angle DK'M = 180^\circ - \angle DAO = 180^\circ - \angle DNM$, and hence $\angle DK'M + \angle DNM = 180^\circ$, so the point $K'$ lies on the circumcircle $\Delta DNM$ – $\omega$, which proves the collinearity of points $G, K, Z$ and $W$. In particular points $G, K$ and $W$ lie on the same line. Let $P$ be a point symmetric to $M$ with respect to $HO$. Then, since $A$ and $W$ are also symmetric with respect to $HO$, then $APMW$ is an isosceles trapezoid, so the point of intersection of its diagonals $PW \cap AM$ lies on $HO$. However $G = HO \cap AM$, so $T$ lies on $GW$, which passes through $K$. So $S$ is the second intersection of $KE$ and the circle $\omega$. Then $\angle KSM = \angle KTM$, and from the isosceles trapezoid $APMW$, $\angle KTM = \angle EAM$, so $\angle EAM = \angle ESM$, from which points $A, S, E$ and $M$ lie on the same circle, but $ES = EM$, from which $AE$ is the bisector of $\angle SAM$, that is $AS$ is the symmedian $\Delta ABC$, which was the proof.