There always exists a circle that touches three of the four sides. For example, the circle that touches $BC$, $CD$ and $DA$ is either the incircle or an excircle of the triangle that has $CD$ as one side and two other sides which are contained in the lines $BC$ and $DA$.

Let $X$, $Y$ and $Z$ be the points where this circle touches $BC$, $CD$ and $DA$, respectively. Suppose $AB$ does not intersect the circle (the case with intersection is similar). Through $B$ there are two tangents to the circle, one of them is $BC$. Let the second tangent touch the circle at $W$ and meet $DA$ at $E$, which lies between $A$ and $D$ because we have assumed that $AB$ does not intersect the circle.

Because the two tangents from a point to a circle have equal length, we have $|EZ| = |EW|$, $|BW| = |BX|$, $|CX| = |CY|$ and $|DY| = |DZ|$. This implies $|EB| + |CD| = |DE| + |BC|$. But we were given $|AB| + |CD| = |AD| + |BC|$, hence $|AB| - |EB| = |AD| - |DE| = |AE|$. This is impossible unless $A = E$. Hence, the circle touches the side $AB$ as well.

If $ABCD$ is a rhombus, such a circle exists. Otherwise, there are two sides of unequal length that start at the same point. We label the vertices of the quadrilateral in such a way that $|AB| < |AD|$. We then have $|CD| > |BC|$. Therefore, we find points $P$ and $Q$ on the line segments $AD$ and $CD$, respectively, such that $|AB| = |AP|$ and $|CB| = |CQ|$. From the given $|AB| + |CD| = |AD| + |BC|$ we obtain $|DP| = |DQ|$ as well.

Connecting $B$, $P$ and $Q$ we obtain three equilateral triangles: $\triangle ABP$, $\triangle CQB$ and $\triangle DPQ$, hence the perpendicular bisectors of $PB$, $BQ$ and $QP$ are the angle bisectors of $\angle PAB$, $\angle BCQ$ and $\angle QDP$, respectively. These three perpendicular bisectors meet at the circumcentre $O$ of triangle $BQP$. Since $O$ is on the three angle bisectors mentioned above, it has the same distance to all four sides of quadrilateral $ABCD$ and so is the centre of a circle that touches all four sides of $ABCD$.