Let $E$ be a point on ray $AD$ beyond $D$ Since $AB$ is tangent to circle $(BFC)$ $\implies$ $\angle BFC = 180^{\circ} - \angle B$ and since $AD$ is tangent to circle $(DFC)$ $\implies$ $\angle DFC = \angle EDC = \angle ADB = \angle B$ Which means that $\angle BFD = \angle BFC - \angle DFC = 180^{\circ} - 2 \angle B = \angle BAD$ $\implies$ $A,F,D,B$ is cyclic. Hence, $\angle AFB + \angle BFC = \angle ADB + \angle BFC = \angle B + 180^{\circ} - \angle B = 180^{\circ}$, giving the required collinearity.