1. Define Projections: Denote by $B_1$ and $B_2$ the orthogonal projections of $B$ on the lines $IA$ and $IC$, respectively. Similarly, let $D_1$ and $D_2$ be the projections of $D$ on the same lines $IA$ and $IC$, respectively.

2. Lemma Application: Consider the following lemma:
Lemma: Let $\triangle ABC$ be a triangle, and denote by $M$ the midpoint of segment $BC$. If $X$ and $Y$ are the orthogonal projections of the vertices $B$ and $C$ on the internal angle bisector of angle $BAC$, then $MX = MY = \frac{|b - c|}{2}$.

Proof of Lemma:
- Let $B'$ be the intersection of the line $BX$ with the sideline $CA$. Since $\triangle ABB'$ is isosceles, the length of segment $CB'$ is $|b - c|$.
- $X$ and $M$ are the midpoints of segments $BB'$ and $BC$, respectively. Thus, $XM = \frac{CB'}{2} = \frac{|b - c|}{2}$.

3. Apply Lemma to Quadrilateral: Returning to the problem, let $T$ be the midpoint of the diagonal $BD$. According to the lemma applied to $\triangle BAD$, we have:
$$TB_1 = TD_1 = \frac{|AB - AD|}{2}$$
Similarly, applying the lemma to $\triangle BCD$, we have:
$$TB_2 = TD_2 = \frac{|BC - CD|}{2}$$

4. Use Pitot's Theorem: For a circumscribed quadrilateral $ABCD$, Pitot's theorem states:
$$AB + CD = AD + BC$$
This implies:
$$\frac{|AB - AD|}{2} = \frac{|BC - CD|}{2}$$

5. Conclude Equal Distances: From the above, we conclude:
$$TB_1 = TB_2 = TD_1 = TD_2$$
This means that the projections of $B$ and $D$ on the lines $IA$ and $IC$ lie on a single circle with center $T$, the midpoint of diagonal $BD$.

$\blacksquare$