Prove that a circle can be inscribed in the hexagon Given a convex hexagon $ABCDEF$. All its sides are equal (it can be irregular). Furthermore, $AD = BE = CF$. How can I prove that a circle can be inscribed in this hexagon?

 A: Let $O$ be the intersection point of diagonals $AD$ and $BE$ (see diagram below) and set $a=OB$, $b=OD$, $d=AD=AE$, $\theta=\angle AOB=\angle DOE$. Then by the cosine rule we have:
$$
a^2+(d-b)^2-2a(d-b)\cos\theta=b^2+(d-a)^2-2b(d-a)\cos\theta,
$$
which simplified gives:
$$
(b-a)(1-\cos\theta)=0,
\quad\hbox{that is:}\quad
a=b.
$$
It follows that $OBC\cong ODC$ and $OAF\cong OEF$, so that 
$\angle COB\cong \angle COD\cong \angle AOF\cong (\pi-\theta)/2$
and points $COF$ are aligned.
By repeating the same argument given above for diagonals $AD$ and $CF$ we get  $OF=a$, $OC=d-a$. It follows that all six triangles in the diagram are congruent between them and in particular they all have the same altitude from their common vertex $O$. A circle of center $O$ and having that common altitude as radius will therefore touch all six red sides of the hexagon.
Notice that if $a\ne d/2$ the hexagon is NOT regular. All six angles with vertex at $O$ are however congruent and they thus measure 60°.

EDIT.
One could prove that $a=b$ even without trigonometry. Triangles $ABO$ and $EDO$ have congruent between them the angles of vertex $O$ and the sides opposite to them. In addition, the other two sides of each triangle have the same difference.
It is well known that if two triangles have the same base $PQ$ and the angles opposite to $PQ$ both congruent, then the third vertex of both triangles lies on the arc of a well defined circle having $PQ$ as a chord. 
Moreover, if the difference of the distances of the third vertex from $P$ and $Q$ is the same for both triangles, then the third vertex must also belong to a well defined hyperbola having $P$ and $Q$ as foci. This third vertex is then one of the two intersections between circle arc and hyperbola, and both possible triangles are congruent between them.
A: Lemma. (1) Diagonal $BE$ is the angle bisector of both angles $\angle \, ABC$ and $\angle \, DEF$; 
(2) Diagonal $CF$ is the angle bisector of both angles $\angle \, BCD$ and $\angle \, EFA$; 
(3) Diagonal $AD$ is the angle bisector of both angles $\angle \, FAB$ and $\angle \, CDE$;
Finally, the three diagonals $AD, \, BE$ and $CF$ intersect at a common point I.
Proof. Look at quadrilateral $ACDF$. Reflect point $A$ in the line $DF$ and let $A^*$ be the symmetric image of $A$ with respect to $DF$. Then $$ \angle \, DA^*F = \angle \, DAF $$ as well as $FA^* = FA = CD$ and $A^*D = AD = CF$. Since $A^*D = CF$ and $FA^*=CD$ the quad $FCDA^*$ is a parallelogram and thus $$\angle \,DCF = \angle \, DA^*F =  \angle \, DAF$$ Therefore quad $ACDF$ is inscribed in a circle and since $FA = CD$, the quad $ACDF$ is in fact an isosceles trapezoid with $AC$ parallel to $DF$. For that reason, the parallel segments $AC$ and $DF$ have a common orthogonal bisector and since $BA = BC$ and $ED =EF$, the points $B$ and $E$ must lie on that orthogonal bisector, i.e. the line $EF$ is the orthogonal bisector of segments $AC$ and $DF$ simultaneously. However, since triangles $ABC$ and $DEF$ are isosceles, the orthogonal bisector $BE$ is at the same time the angle bisector of both angles $\angle \, ABC$ and $\angle \, DEF$. Observe that since the line $BE$ is the orthogonal bisector of the two parallel sides $AC$ and $DF$ of the isosceles trapezoid $ACDF$ its two diagonals $AD$ and $CF$ intersect in a common point which lies on $BE$. The rest of the lemma follows analogously.
$$ $$
Completing the proof: Observe that by the Lemma, the lines $AD, \, BE,$ and $CF$ all meet at a common point, we denote by $I$. Consequently, by the Lemma, $I$ is the common intersection point of all angle bisectors of the angles at the vertices of the hexagon $ABCDEF$. Therefore, there is a circle inscribed in the hexagon with point $I$ the incenter. As a consequence we get that $IA=IC=IE$ and $IB = ID=IF$.
