Solution. Consider the problem as solved. Let the given vertices of the quadrilateral be denoted by $A, B, C$, and the fourth vertex by $D$. Choose the labeling such that $A B \geqq B C$ holds.

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Figure 1

Since the quadrilateral is a cyclic quadrilateral, the point $D$ lies on the circle determined by the points $A, B, C$. From the property of the tangential quadrilateral, it follows that $A B + C D = B C + A D$, or rearranging, $A B - B C = A D - C D = d$. Measure the distance $d$ from $A$ along the segment $A D$. Denote the endpoint of the resulting segment by $K$. Then the triangle $C K D$ is isosceles, and the angle at vertex $D$ is known, as it complements the angle $A B C = \beta$ to $180^{\circ}$. Thus, $D K C \angle = \frac{180^{\circ} - (180^{\circ} - \beta)}{2} = \frac{\beta}{2}$, and therefore $A K C \angle = 180^{\circ} - \frac{\beta}{2}$; hence, in triangle $A K C$, we know two sides and one angle.

Based on this, the construction of the quadrilateral is as follows: draw a circle around the points $A, B, C$, then draw a circle with a viewing angle of $180^{\circ} - \frac{\beta}{2}$ over the side $A C$. Cutting this with a distance of $d = A B - B C$ from $A$ gives the point $K$. Finally, the line $A K$ intersects the circle at $D$.

The quadrilateral constructed in this way is clearly a cyclic quadrilateral and also a tangential quadrilateral. Since $C K D \angle = 180^{\circ} - A K C \angle = \frac{\beta}{2}$, $K D C \angle = 180^{\circ} - \beta$, so $K D C$ is an isosceles triangle, $C D = K D$, and thus $A B + C D = (B C + d) + C D = B C + (d + K D) = B C + A D$.

If the points $A, B, C$ lie on a straight line, then the problem has no solution. If $A, B, C$ do not lie on a straight line, then there are always 3 solutions; in the solution, we assumed that the order of the vertices of the quadrilateral is $A, B, C, D$, i.e., $D$ lies on the arc of the circle $A C$ that does not contain $B$. However, if only 3 points are given, $D$ can lie on any of the 3 arcs determined by them (Figure 2). If $d = 0$ (in any case), then the cyclic quadrilateral is a kite, and $D$ is cut out from the circle by the perpendicular from $B$ to $A C$.

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Figure 2