Denote by $ B_{1}$, $ B_{2}$ the orthogonal projections of $ B$ on the lines $ IA$, and $ IC$, respectively, and similarly, let $ D_{1}$, $ D_{2}$ be the projections of $ D$ on the same lines $ IA$, and $ IC$, respectively. We consider the following preliminary result: Lemma. Let $ ABC$ be a triangle, and denote by $ M$ the midpoint of segment $ BC$. If $ X$, $ Y$ are the orthogonal projections of the vertices $ B$, $ C$ on the internal angle bisector of angle $ BAC$, then $ MX = MY = |b - c|/2$. Proof. We shall resume to proving that $ MX = |b - c|/2$. For this, let $ B^{\prime}$ be the intersection of the line $ BX$ with the sideline $ CA$. Since the triangle $ ABB^{\prime}$ is isosceles, the length of segment $ CB^{\prime}$ is $ |b - c|$. On other hand, $ X$, $ M$ are the midpoints of segments $ BB^{\prime}$, and $ BC$, respectively. Thus $ XM = CB^{\prime}/2 = |b - c|/2$. This proves our Lemma. Returning to the problem, let $ T$ be the midpoint of the diagonal $ BD$. According to the Lemma, applied for the triangle $ BAD$, we have that $ TB_{1} = TD_{1} = |AB - AD|/2$. Similarly, according to the same Lemma, this time applied in triangle $ BCD$, we have that $ TB_{2} = TD_{2} = |BC - CD|/2$. On the other hand, the quadrilateral $ ABCD$ beeing circumscribed, $ AB + CD = AD + BC$ (Pithot's theorem), and hereby, we conclude that $ TB_{1} = TB_{2} = TD_{1} = TD_{2}$, i.e. the projections of $ B$, and $ D$ on the lines $ IA$, and $ IC$ lie on a single circle (with center $ T$, the midpoint of diagonal $ BD$).

I think it is easy to calculate angles of triangle IB1B2 an ID1D2 to get them similar which gives result

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