a) Consider cases of the location of the vertex angle of the triangle $BCF$ (Fig. 27).
If $F$ were the apex, then $F$ would lie on the perpendicular bisector of the side $BC$, i.e., on the line $AD$, whence $F = A$. Therefore $AC$ would be a tangent of the incircle of the triangle $ABD$. But the lines $AB$ and $AD$ are tangents of the same circle. There can be at most two tangents drawn from one point to one circle. Hence this case is impossible.
Let $B$ be the apex. As $CBF$ is a base angle of the isosceles triangle $ABC$, we have $\angle CBF < 90^\circ$. Hence $\angle BCF > 45^\circ$. Let $K$ be the tangent point of the line $CF$ and the incircle of the triangle $ABD$ and let $L$ be the projection of the point $K$ onto the line $BC$. By construction, $KL < 2r$, where $r$ is the radius of the incircle

Fig. 27

of the triangle $ABD$. On the other hand, $\angle LCK = \angle BCF > 45^\circ$ implies $KL > CL > CD = BD > 2r$. Hence this case is impossible, too.
This shows that the apex of triangle $BCF$ can only be $C$.

b) Let the apex be $C$. Fix a point $D$, mutually perpendicular lines $l_1$ and $l_2$ both passing through $C$, and circle $c$ that is tangent to both lines; let the radius of the circle be $r$. Choose point $A$ on $l_1$ in such a way that $DA > 2r$ and the tangent point of line $l_1$ and circle $c$ lies on the line segment $DA$. Point $B$ is determined by the location of $A$ as the point of intersection of line $l_2$ and the second tangent line of circle $c$ passing through $A$, point $C$ is symmetric to $B$ w.r.t. $DA$ and $F$ is defined as in the problem. When point $A$ moves away from $D$, points $B$ and $C$ get closer to $D$, whence $\angle BAC$ decreases and $\angle BCF$ increases. Thus these angles can equal only in one case.