In my opinion this problem is a bit too hard for the Problem 4 position but it's still really nice. There is such a lot going on in this picture. Let $J = \overline{BL} \cap \overline{AC}$. Let $N$ be the midpoint of arc $BAC$. We first make a few minor observations. Claim : Quadrilateral $BOCL$ is a rhombus and quadrilateral $OLJK$ is an isosceles trapezoid. Proof : Note that, \[2\measuredangle OBC = \measuredangle BOC = 2\measuredangle BAC \]Thus, $\measuredangle OBC = 90 - \measuredangle BAC = 30^\circ$. But clearly, \[\measuredangle CBL = \measuredangle CAL = 30^\circ \]So, $\measuredangle OBC = \measuredangle CBL$. Now, this means that $L$ is the reflection of $O$ across the midpoint of $BC$, since it is well known that $OL$ is the perpendicular bisector of $BC$. From this, the result follows. Further, it is now easy to see that $OLJK$ is an isosceles trapezoid by symmetry since $L$ is the reflection of $O$ across $BC$ and $J$ is the reflection of $K$ across $BC$. Now, it is not hard to see that points $B$ , $O$ and $K$ are collinear since \[\measuredangle KBC = \measuredangle CBJ = \measuredangle CBL = 30^\circ = \measuredangle OBC\]We can the prove our first key claim. Claim : Points $N$ , $A$ and $K$ lie on a straight line. Proof : Note that in triangles, $\triangle LCJ$ and $\triangle KON$, \[\measuredangle NOK = \measuredangle LOB = 60^\circ = \measuredangle CAB = \measuredangle CLJ\]Further, we have that $CL = OC = ON$ since $BOCL$ is a rhombus and $LJ=OK$ since $OLJK$ is an isosceles trapezoid. Thus, it follows that $\triangle LCJ \cong \triangle KON$. Now, as a result of that we note, \[\measuredangle KNO = \measuredangle JCL = \measuredangle ANL = \measuredangle ANO \]from which the claim clearly follows. Now, we are left to deal with $D$. We now define $D'$ to be the reflection of the intersection $E$ of $AB$ and $CL$ across $BC$. Then, $D'$ satisfies all the properties we proved above for $K$ as well. In particular, $D'$ lies on lines $\overline{CO}$ and $\overline{AN}$. Now, we can also note that by Brokard's Theorem on cyclic quadrilateral $ABLC$ we have that $O$ is the orthocenter of $\triangle EJM$. This means that $M$ is the orthocenter of $\triangle OEJ$. Reflecting across the line $BC$, we have that $M$ is the orthocenter of $\triangle LD'K$. Thus, $D'M \perp LK$, which means that $D' = D$. Thus, points $D$ and $K$ both lie on the external angle bisector of $\angle BAC$ which passes through $A$ and we are done.

Moving points kills. You can also notice that line $AK$ moves linearly as you move $A$ along the circumcircle. This means it must pass through a fixed point, and we see immediately that this point is the midpoint of arc $BAC$. Now you need just 3 cases to finish which is even more trivial.

Very neat problem! Here is a sketch of a solution using a lil inversion: Let $A'$ be the reflection of $A$ across $BC$ and let $D'=CO\cap A'B$. We will show that $D=D'$. Also notice that $K$ can be redefined as $BO\cap A'C$. Observe that angles $OKC$ and $OAC$ are both $\angle BCA-30$, and thus $\square OACK$ is cyclic. By a similar argument, so is $BOAD'$. Thus, $\angle KAC=\angle KOC=60\implies \angle KAL=90$. Similarly, $\angle D'AL=90$ and thus $K-A-D'$ colinear. Now, let $X\in LK$ such that $MX\perp LK$ and similarly $Y\in LD'$ such that $MY\perp LD'$. We claim that the following three are circles. $\square MYXL$ (by construction) $\square BOMY$ (invert circle $BOAD'$ with respect to the circle with $L$ as the center and $LB$ as the radius) $\square BOXL$ (invert line $B-O-K$ with respect to the same circle as earlier) By radical axis, $BO$, $MY$ and $XL$ are concurrent, which tells us that $K-M-Y$ are colinear. Further, $KM\perp D'L$. Or similarly, $D'M\perp KL$. This tells us that $D'=D$ and thus our proof is complete.