There are always 3 such points D, D', D'' on the circumcircle $K$ f the triangle $\triangle ABC$, depending on which triangle side is taken as one of the diagonals of the bicentric quadrilateral with the vertices A, B, C, D (or D' or D''). WLOG, assume that CA is selected to be a diagonal, so that the point D lies on the circumcircle arc CA opposite to the vertex B . Let I be the incenter of the bicentric quadrilateral ABCD, which lies on the known bisector BI of the angle $\angle ABC$ and on the unknown bisectors of the angles $\angle BCD, \angle DAB$. The angle $\angle CIA = 360^\circ - (\angle ICD + \angle CDA + \angle IAD) = 360^\circ - \left(\frac{\angle BCD}{2} + \angle CDA + \frac{\angle DAB}{2}\right)$ Since the quadrilateral ABCD is also cyclic, $\angle BCD + \angle DAB = \angle ABC + \angle CDA = 180^\circ$. Substituting this to the above equation, $\angle CIA = 360^\circ - (90^\circ + 180^\circ - \angle ABC) = \angle ABC + 90^\circ$ First, assume that the angle $\angle ABC < 90^\circ$ is acute. Then $\angle CIA = \angle ABC + 90^\circ < 180^\circ$ and we have to construct a circle (E) passing through the points C, A, such that the angles spanning its minor arc CA on the same side of the line CA as the vertex B are equal to the known angle $\angle CIA$. The incenter of the quadrilateral ABCD lies on this arc. To do this, we draw tangents to the circumcircle $K$ at the vertices C, A. Since we assume the angle $\angle ABC$ to be acute, these tangents intersect at a point E on the opposite side of the line CA than the vertex B. The angles $\angle CAE = \angle ACE = \angle ABC$, hence, the angle $\angle CEA = 180^\circ - 2 \angle ABC$ and the angles spanning the minor arc CA if the circle (E) centered at E and passing through C, A are equal to $180^\circ - \frac{180^\circ - 2\angle ABC}{2} = \angle ABC + 90^\circ$ The incenter I of the bicentric quadrilateral ABCD is the intersection of the bisector of the angle $\angle ABC$ with the minor arc CA of the circle (E). Once we have the incenter I, we can construct the circle (I) tangent to the lines BA, BC and then draw the other tangents to (I) from the vertices C, A intersecting at the desiren point $D \in K$. If the angle $\angle ABC = 90^\circ$ is right, the circle (E) simply degenerates into the line CA (which is now a diameter of the circumcircle K) and the incenter I of the bicentric quadrilateral ABCD is the foot of the bisector of the right angle $\angle ABC$. Second, assume that the angle $\angle ABC > 90^\circ$ is obtuse. Then $\angle CIA = \angle ABC + 90^\circ > 180^\circ$ and we have to construct a circle (E) passing through the points C, A, such that the angles spanning its minor arc CA on the opposite side of the line CA thans the vertex B are equal to the known angle $180^\circ - \angle CIA < 180^\circ$. The construction of the circle (E) is exactly the same as before.

Here is another construction: We will use the theorem: A convex quadrilateral ABCD is tangential iff AB+CD = BC+DA (see http://www.mathlinks.ro/Forum/viewtopic.php?t=58984, see also the nice proof from nttu) So, it's enough to find a point $D$ on the arc $AC$ (not containing $B$), such that $AB+CD = BC+DA$ (similarly for the points $D', D''$ specified by yetti) If $AB=BC$, then the problem is easy. We bring the bisector of $\angle{ABC}$, which intersects the circle at a point $D$. Then $DA=DC \Rightarrow AB+CD=BC+DA \Rightarrow ABCD$ is tangential. The incenter $I$ of $ABCD$ can be found on the intersection of $BD$ with the angle bisecor of $\angle{DAB}$. Suppose that $AB>BC$ (I use nttu's proof) $AB+CD = BC+DA \Rightarrow DA>CD$ Let $M \in BA$, such that $BM = BC$ and $N \in DA$, such that $DN = DC$ $AB+CD = BC+DA \Rightarrow$ $AM+MB+CD = BC+DN+AN \Rightarrow$ $AM+BC+CD = BC+CD+AN \Rightarrow AM=AN$ Let $P$ be the midpoint of the arc $AC$ (the arc whics contains the point $B$) The triangles $PDN,PDC$ are equal Indeed: $PD=PD$ $DN=DC$ $\widehat{PDN} = \widehat{PDC}$, because $DP$ is the angle bisector of $\widehat{ADC}$ As a result, $PN=PC$ Construction We construct the points: the midpoint $P$ of the arc $AC$ which contains the point $B$ $M \in BA$, such that $BM = BC$ the intersection $N$ of the circles $(A,AM)$ and $(P,PA)$ ; from the two points of intersection we choose the one which is in the interior of the circle $K$. The ray $AN$ intersects again the circle $K$ at a point $D$ The triangles $PND,PCD$ are equal, because $\widehat{NDP} = \widehat{CDP}$ $PD=PD$ $\widehat{PND} = 180^0- \widehat{ANP} = 180^0- \widehat{PAN} = \widehat{PCD}$ Hence $DN=DC$ $ \begin{array}{cc} AB+CD = BM+AM+CD \\ BC+AD = BC+AN+DN \end{array}\}\Rightarrow AB+CD=BC+AD \Rightarrow ABCD$ is tangential We bring the angle bisector of the angle $\widehat{BAD}$, and where it intersects the segment $PD$, there is the incenter $I$.