Brazilian Mathematical Olympiad

If $ABCD$ is a rhombus the incircle touch the respective triangle in the midpoint of the diagonals, which belongs to all four circles.

Now suppose the circles have a common point $P$. Since the incircle are contained in its triangle and the intersection of the triangles $ABC$, $BCD$, $CDA$, $DAB$ is the intersection of the diagonals, $P$ must coincide with $O$.

Let's prove that the diagonals are perpendicular. Suppose that it's not the case. So we may suppose without loss of generality that $\angle AOB = \angle COD > 90^\circ$. Let $l_1, l_2, l_3$ and $l_4$ be the incenters of $ABC$, $BCD$, $CDA$ and $DAB$, respectively. Thus $l_1$ and $l_4$ are in the interior of $\angle ABC$ and $l_2$ and $l_3$ are in the interior of $\angle COD$.

Let $E$ and $F$ be the intersection of $AI_4$, $BI_1$ and $CI_2$, $DI_3$, respectively. So $\angle BEA > \angle AOB > 90^\circ \Rightarrow 180^\circ - \frac{\angle DAB}{2} - \frac{\angle ABC}{2} > 90^\circ \Leftrightarrow \angle DAB + \angle ABC < 180^\circ$. Using a similar argument on $F$ we obtain $\angle BCD + \angle CDA < 180^\circ$. Summing these two inequalities, we obtain that the sum of the internal angles of the quadrilateral $ABCD$ is less than $360^\circ$, contradiction. So $AC$ and $BD$ are perpendicular and, moreover, the four incenters $l_1, l_2, l_3$ and $l_4$ lie in the diagonals. This means that the bisectors and altitudes from vertices $A, B, C, D$ with respect to the triangles $DAB$, $ABC$, $BCD$, $CDA$ coincide and hence these triangles are isosceles. Thus $DA = AB = BC = CD$ and $ABCD$ is a rhombus.