Assume that the angle bisectors of the angles $\angle B$ and $\angle D$ intersect at $P$ (Fig. 1). We shall prove that the hexagon $ABCDEF$ has an inscribed circle, whose center is $P$. Then the conclusion follows from Brianchon's Theorem.
The equality $AB = BC$ implies that the triangles $ABP$ and $CBP$ are congruent. Hence we have $\angle BAP = \angle BCP = x$. Similarly, triangles $CDP$ and $EDP$ are congruent, so we obtain $\angle DCP = \angle DEP = y$.

Fig. 1

Moreover, we have $AP = CP = EP$, which together with the equality $AF = EF$ implies that the triangles $AFP$ and $EFP$ are congruent. Thus the angle bisector of the angle $\angle F$ passes through the point $P$ and $\angle FAP = \angle FEP = z$.
Now the equalities $\angle A = \angle C = \angle E$ are equivalent to $z+x=x+y=y+z$, which yields $x=y=z$. Therefore the angle bisectors of the angles $\angle A$, $\angle C$ and $\angle E$ all pass through the point $P$. Thus $P$ is the center of the inscribed circle of the hexagon $ABCDEF$, as claimed.