Problem 1105

Let's use the notation from the figure.

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Let the center of the circle $k$ passing through points $A, B, C$ be $O$. It is known that if $A B \neq B C$, the angle bisector of $\angle A B C$ and the perpendicular bisector of side $A C$ have exactly one common point, and this common point lies on the circumcircle of triangle $A B C$. This common point is the point $E$, which lies on the perpendicular bisector of side $A C$.

Let the intersection of the perpendicular bisector of $O E$ and side $A C$ (i.e., the midpoint of $A C$) be denoted by $M$. Since $\angle E M D = 90^\circ$, by Thales' theorem, $M$ lies on the circumcircle of triangle $D F E$. The line $E O$ intersects the circle $k$ again at point $H$. Since $\angle D F E = 90^\circ$ and $\angle H F E = 90^\circ$ (by Thales' theorem), the line $D F$ passes through $H$.

By Thales' theorem, $\angle H B E = 90^\circ$, since $H E$ is a diameter. Therefore, since $\angle H B D + \angle D M H = 90^\circ + 90^\circ = 180^\circ$, the quadrilateral $H B D M$ is a cyclic quadrilateral. Thus, by the theorem of inscribed and central angles, $\angle M H D = \angle M B D$.

Since $H B F E$ is also a cyclic quadrilateral, by the theorem of inscribed and central angles, $\angle E H F = \angle E B F$.

Therefore, $\angle D B F = \angle M B D$, which means that the reflection of $B F$ over $B D$ is $B M$, i.e., $B F$ is indeed the median.

Several people noted that the statement of the problem is equivalent to the fact that the line $B F$ is a symmedian of triangle $A B C$. For more details on the symmedian, see László Surányi: On the less well-known notable points of the triangle, Part II (KöMal-1984/November).