Let triangle $ABC$, such that $\measuredangle A=120^{\circ}$ and $AB\not=AC$. $AL -$ bisector, $AK -$ median. The point $O -$ center of the circle circumscribed around the triangle, $OL\cap AK= F$. Find $\measuredangle BFC$.
I made a drawing. I have a hypothesis that $\measuredangle BFC=60^{\circ}$. But I can not prove it.
On
Alternative solution.
Let $c$ be the circle circumscribed around triangle $ABC$ and let $c'$ be the image of $c$ with respect of the reflection in the line $BC$. Then due to the fact that $\measuredangle \, BAC = 120^{\circ}$ the circle $c'$ goes through the point $O$, which is the center of $c$.
Draw the orthogonal bisector $OK$ of edge $BC$ and let $O'$ and $M$ be the two intersection points of $OK$ with the circle $c$ so that $O'$ and $A$ are on the same side of $BC$. Then points $A, L, M$ are collinear because $AL$ is the interior angle bisector of $\angle \, BAC$.
Draw the line $AO'$ and let $L^*$ be the intersection point of $AO'$ with $BC$. Then line $O'A \equiv AL^*$ is the external bisector of angle $\angle \, BAC$, the two lines $AL \equiv AM$ and $O'L^* \equiv AL^*$ are orthogonal and the circle $c^*$ defined by the points $LAL^*$, also known as the Apollonius' circle, is centered at the midpoint of segment $LL^*$ and is orthogonal to all circles that pass through the points $B$ and $C$. In particular, $c^*$ is orthogonal to both circles $c$ and $c'$.
Next, draw circle $c_0$ centered at the point $K$ and of radius $KB = KC$. Then $c_0$ is also orthogonal to $c^*$. Observe that $KO \cdot KM = KO'\cdot KM = BK^2$ so if you perform an inversion with respect to $c_0$, point $M$ is mapped to point $O$ and hence circle $c$ gets mapped to circle $c'$, while $c^*$ is mapped to itself (due to orthogonality with $c_0$).
Let $F'$ be the inverse image of $A$ with respect to $c_0$. Point $F'$ lies on $c'$ because its preimage $A$ lies on $c$ ($c$ is the preimage of $c'$).Point $F'$ also lies on $c^*$ because $A$ lies on $c^*$ and $c^*$ is mapped to itself when inverted in $c_0$. Then the three points $K, A, F'$ are collinear and $F' \in c' \cap c^*$.
Consider the quad $L^*ALM$. It is inscribed in a circle because $\measuredangle \, MAL' = 90^{\circ} = \measuredangle \, MKL^*$. Now, under the inversion with respect to $c_0$, point $L^*$ is mapped to point $L$, point $A$ is mapped to point $F'$, point $M$ is mapped to $O$ and point $K$ is mapped to infinity, so the circle defined by $L^*ALM$ is mapped to the straight line $OL \equiv OF'$ meaning that the three points $F', L, O$ are collinear. Thus, point $F'$ is the intersection point of $AK$ and $LO$ which was initially denoted by $F$. Thus $F' \equiv F$.
Consequently, as $F$ belongs to the circle $c'$, angle $\measuredangle \, BFC = 60^{\circ}$.
On
Here's one more elementary solution using only angle-chasing.
First let's somewhat restate the problem. Denote the intersection of $OL$ with $\odot ABC$ with $D$ and $G$, s.t. $D$ is one the minor arc $BC$. Also WLOG let's assume $AB<AC$, so actually $D$ lies on the arc $AB$. Now reflect $BC$ over $BD$ and again over $CD$. Let the two reflections concur at $F'$. We'll prove that $F \equiv F'$. For this it's enough to prove that $F$ lies on $OL$ and $AK$. From now on for simplicity let $\measuredangle DBC = \beta, \measuredangle DCB = \gamma$.
Obviously now $D$ is the incenter of $\triangle F'BC$. So we have that $\measuredangle F'DB = 90^{\circ} + \gamma$. Now we have that $\measuredangle BDG = \measuredangle BCG = \measuredangle DCG - \measuredangle DCB = 90^{\circ} - \gamma.$ So therefore: $\measuredangle F'DB + \measuredangle BDG = 90^{\circ} + \gamma + 90^{\circ} - \gamma = 180^{\circ}$. Hence $F'$ lies on $DG \equiv OL$.
Now obviously we have that $\measuredangle BF'C = 60^{\circ}$ and from Angle Bisector Theorem we have that $\frac{F'B}{F'C} = \frac{BL}{LC} = \frac{AB}{AC}$. So now as $K$ is the midpoint of $BC$ we have that $[F'KB] = [F'KC] \implies BF' \cdot \sin \angle KF'B = CF' \cdot \sin \angle KF'C$. Now from the Sine Theorem on $\triangle ABC$ we have that:
$$\frac{\sin \angle ABC}{\sin \angle ACB} = \frac{AC}{AB} = \frac{F'C}{F'B} = \frac{\sin \angle KF'B}{\sin \angle KF'C}$$
But as $\measuredangle ABC + \measuredangle ACB = \measuredangle KF'C + \measuredangle KF'A = 60^{\circ}$ we must have $\measuredangle KF'C = \measuredangle ACB; \measuredangle KF'B = \measuredangle ABC$.
Now similarly for $\triangle ABC$ we have: $[KAB] = [KAC] \implies AB \cdot \sin \angle KAB = AC \cdot \sin \angle KAC$. But similarly as before we have:
$$\frac{\sin \angle KAB}{\sin \angle KAC} = \frac{AC}{AB} = \frac{F'C}{F'B} = \frac{\sin \angle F'BC}{\sin \angle F'CB}$$
And as before we have that $\measuredangle KAB = \measuredangle F'BC; \measuredangle KAC = \measuredangle F'CB$, as $\measuredangle KAB + \measuredangle KAC = \measuredangle F'BC + \measuredangle F'CB = 120^{\circ}$. So finally we have that: $\measuredangle F'KB = 180^{\circ} - \measuredangle KF'B - \measuredangle KBF' = 180^{\circ} - \measuredangle ABC - \measuredangle KAB = \measuredangle AKB$. Therefore $F'$ lies on $AK$.
Therefore as $F'$ lies on $OL$ and $AK$ we have that $F \equiv F'$. So finally $\measuredangle BFC = \measuredangle BF'C = 60^{\circ}$
Let $G$ be the antipode of $O$ in the circumcircle of $OBC$ and $\Gamma$ the circumcircle of $ABC$.
If we consider a circle inversion with respect to $\Gamma$, $G$ is the inverse of $K$ and $A$ is the inverse of itself. It follows that the inverse of the $AK$-line is the circle through $O,A,G$, and $F$ it the inverse of the intersection of the $OL$-line with the circle through $O,A,G$. If we prove that $OLAG$ is a cyclic quadrilateral, we get that $L$ is the inverse of $F$, hence $OBFC$ is a cyclic quadrilateral and $\widehat{BFC}=180^\circ-\widehat{BOC}=\color{red}{60^\circ}$ as claimed.
In order to prove that $O,L,A,G$ lie on the same circle, it is enough to consider $H$ as the symmetric of $G$ with respect to $BC$, notice that both $HBC$ and $GBC$ are equilateral triangle and check that $$ HL\cdot HA = HO\cdot HG. $$ In particular, neither $HL\cdot HA$ or $HO\cdot HG$ really depend on the position of $A$ on the $BC$-arc. Hint for proving such fact: