Let $ O$ be the center of the circumcircle. From $ OI = \sqrt {R + r^2 - r\sqrt {4R^2 + r^2}}$, we can construct the point $ I$ - the incenter. If $ P$ is the intersection of the diagonals, then $ OP = \dfrac{2R^{2}d}{R^{2} + d^{2}}$ (proved here}). So, the point $ P$ is also constructible. Points $ P$, $ I$, $ O$ lie on a line in this order. If $ K$ and $ L$ are midpoints of the diagonals $ AC$ and $ BD$, then $ OL \perp BD$ and $ OK \perp AC$, and we deduce, that quadrilateral $ OLKP$ is inscribed in a circle $ \omega$ with diameter $ PO$ (this circle can be constructed). Points $ K$, $ L$ and $ I$ lie on the Newton's line. We need to construct a line through $ I$, which cuts $ \omega$ at points $ K$ and $ L$, such that $ \angle{KPL} = \alpha$, where $ \alpha$ - the given angle betwwen diagonals. If $ M$ is midpoint of $ PO$, then $ \angle{KML} = 2\alpha$, and our problem is to construct the isosceles triangle $ MKL$ with known vertex $ M$, angle $ \angle{KML}$, length of the side $ KM$ and point $ I$ on the side $ KL$. Apply Stewart in $ \triangle{KML}$ for $ MI$ and get $ \small{MI^2 = \dfrac{MK^2\cdot IL + ML^2\cdot IK - IK\cdot IL \cdot KL}{KL} = MK^2 - IK\cdot IL = MK^2 - IK(KL - IK) = }$ $ \small{ = MK^2 + IK^2 - IK\cdot KL \Longleftrightarrow IK^2 - IK\cdot KL - MI^2 = 0}$ We solve the quadratic $ IK$. Note that $ KL = 2MK\sin \alpha$. Because the construction is possible, the equation has two roots. But just one satisfies us because $ x_1\cdot x_2 = - MI^2 < 0$. So we can find $ IK$. We construct the circle $ \Omega$ with the center $ I$ and radius $ IK$. Clearly, circles $ \omega$ and $ \Omega$ intersect at $ K$ (there exist two such $ K$s, we''l continue with one of them). The line $ KI$ meets $ \omega$ at $ L$. Line $ PL$ meets the circumcircle at $ B$ and $ D$, but line $ PK$ at points $ A$ and $ C$. The construction is finished.

Let $A,B,C,D$ be the vertices of the wanted quadrilateral with incircle $ (I)$ and circumcircle $ (O).$ According to Fuss relation (mentioned by Ahiles), distance $ IO$ is constructible, then we can place the incircle and the circumcircle. According to Poncelet porism, $O,I$ and diagonal intersection $P \equiv AC \cap BD$ are fixed for all quadrilaterals with incircle $(I)$ and circumcircle $(O)$ $\Longrightarrow$ $P$ is constructible. Let $ M,N$ the orthogonal projections of $ O$ on $AC,BD,$ it follows that $ OM$ and $ ON$ are perpendicular bisectors of $AC,AB$ and $ MN \equiv n $ is the Newton line of the quadrangle $ ABCD$ passing through its incenter $I.$ Hence it suffices to draw the line $n$ passing through $ I$ such that $ \angle MPN$ is given. If $K$ is center of the circle with diameter $ OP,$ then central angle $ \angle MKN$ is known as well $\Longrightarrow$ Isosceles triangle $ \triangle MKN$ is constructible. Placing the constructed chord $ MN$ in the circle $ (K)$ passing through $ I$ completes the quadrilateral $ABCD$ easily.

Ok let \(\angle A, \angle B\) be the two adjacent angles of the quadrilateral, draw a \(\Delta ABC\) with the given two angle measures. Let \(A'B'C'\) be the intouch triangle of \(\Delta ABC\), let \(X \in A'B'\)\(: C'X \perp B'A'\), let \(AX \cap BC =D, BX \cap CA =E, \square ACDE\) is the required .