The circumcenter $O$, the incenter $I$, and the midpoint $M$ of a diagonal of a bicentral quadrilateral were marked. After this the quadrilateral was erased. Restore it.

1. **Observations:**
   - Let \( M \) be the midpoint of the diagonal \( AC \) and \( N \) be the midpoint of the diagonal \( BD \).
   - Let the incircle of \( ABCD \) touch the sides \( AB, BC, CD, DA \) at points \( X, Y, Z, W \) respectively.
   - Let \( E \) and \( F \) be the midpoints of \( XZ \) and \( YW \) respectively.
   - Let \( G = AB \cap CD \) and \( H = AD \cap BC \).
   - Denote by \( V \) the intersection of \( AC \) and \( BD \).

   The following results hold:
   - \( XZ \perp YW \)
   - \( V \in XZ \) and \( V \in YW \)
   - \( V \in IO \)
   - \( I \in MN \)
   - \( EF \perp MN \)
   - \( G \in EI \) and \( H \in FI \)
   - \( V \) is the orthocenter of \( \triangle GOH \)

2. **Proof of \( XZ \perp YW \):**
   \[
   180^\circ - \angle YVZ = \angle VYZ + \angle VZY = \angle WYZ + \angle XZY
   \]
   \[
   = \frac{180^\circ - \angle WDZ}{2} + \frac{180^\circ - \angle XBY}{2} = 180^\circ - \frac{\angle ADC + \angle ABC}{2} = 90^\circ
   \]
   \[
   \Longrightarrow \angle YVZ = 90^\circ \Longrightarrow XZ \perp YW
   \]

3. **Construction:**
   - Construct the line \( \ell \) through \( M \), perpendicular to \( OM \). This is line \( AC \) as \( OM \perp AC \), because \( M \) is the midpoint of the chord \( AC \) in \( (ABCD) \), which has center \( O \).
   - Now we know that \( V = OI \cap AC \), so if we intersect \( \ell \) with \( OI \), we'll construct point \( V \).
   - \( ON \perp BD \Longrightarrow ON \perp NV \), but also from the third bullet point in **Observations**, we have that \( I \in MN \), so if we construct the circle with diameter \( OV \) and intersect \( MI \) with this circle for the second time, then that would be point \( N \), so we can construct point \( N \).
   - \( XZ \perp YW \) and \( IE \perp XZ \) and \( IF \perp YW \), so \( IFVE \) is a rectangle, so the midpoint of \( EF \) and \( IV \) are the same point and \( IEVF \) is cyclic.
   - \( EF \perp MN \) from the fifth bullet point, so if we construct \( U \), the midpoint of \( IV \), construct the circle \( k(U, IU) \) and construct the line \( g \) through \( U \), perpendicular to \( MN \), then the intersection points of \( g \) and \( k \) are points \( E \) and \( F \).
   - Therefore, we've constructed \( E \) and \( F \).

4. **Constructing points \( G \) and \( H \):**
   - We know that \( V \) is the orthocenter of \( \triangle GOH \) by the seventh bullet point, so \( G, H \) are uniquely determined on \( \overline{IE} \) and \( \overline{IF} \).
   - We will write this as a lemma, because it is not trivial.

   **Lemma:**
   We can construct points \( G \) and \( H \) with the points \( O, I, V, E, F \).

   **Proof:**
   - We will only need \( E \) and \( F \) to construct the lines \( \overline{IE} \) and \( \overline{IF} \), let's call them \( t_e \) and \( t_f \) respectively.
   - We mentioned that \( G \in IE = t_e \) and \( H \in IF = t_f \).
   - Let \( R = OV \cap GH \) and \( Q = HV \cap OG \).
   - We mentioned that \( V \) is the orthocenter of \( \triangle GOH \), so \( HQ \perp OG \) and \( OR \perp GH \).
   - Let \( OI = x \), \( IV = y \), \( VR = z \), \( \angle VIF = \alpha \) (we mentioned that \( IFVE \) is a rectangle, so \( \angle VIE = 90^\circ - \alpha \)).
   - We have that:
     \[
     RH = \tan \angle RIH \cdot RI = (y + z) \tan \alpha
     \]
     \[
     GR = \tan \angle RIG \cdot RI = (y + z) \tan (90^\circ - \alpha) = \frac{(y + z)}{\tan \alpha}
     \]
     \[
     GH^2 = (GR + RH)^2 = (y + z)^2 \left( \tan^2 \alpha + 2 + \frac{1}{\tan^2 \alpha} \right)
     \]
     \[
     OH^2 = RH^2 + OR^2 = \tan^2 \alpha (y + z)^2 + (x + y + z)^2
     \]
     \[
     GV^2 = GR^2 + RV^2 = \frac{(y + z)^2}{\tan^2 \alpha} + z^2
     \]
     \[
     OV^2 = (x + y)^2
     \]
     \[
     HV \perp OG \Longrightarrow GH^2 + OV^2 = GV^2 + OH^2
     \]
     \[
     \Longrightarrow (y + z)^2 \left( \tan^2 \alpha + 2 + \frac{1}{\tan^2 \alpha} \right) + (x + y)^2 = \frac{(y + z)^2}{\tan^2 \alpha} + z^2 + \tan^2 \alpha (y + z)^2 + (x + y + z)^2
     \]
     \[
     \Longrightarrow 2(y + z)^2 + (x + y)^2 = z^2 + (x + y + z)^2 \Longrightarrow 2y^2 + 2yz = 2zx \Longrightarrow \boxed{z = \frac{y^2}{x - y}}
     \]
   - However, we have constructed points \( O, I, V \), so we know the values of \( x \) and \( y \), we have \( t_e \) and \( t_f \) so we can construct \( R \) by constructing a segment \( VR' \) with length \( \frac{y^2}{x - y} \) on line \( \overline{OI} \), so that \( V \in R'I \).
   - From what we've shown \( R' \equiv R \), so we have constructed \( R \).
   - Now, if we intersect the line through \( R \), perpendicular to \( OR \) with \( t_e \) and \( t_f \), we'll get \( G \) and \( H \) respectively, so we've proven the lemma.

5. **Constructing points \( X, Y, Z, W \):**
   - Having constructed \( G \) and \( H \), construct the circle \( c(G, \sqrt{GE \cdot GI}) \).
   - The intersection of this circle with the line \( VE \equiv XZ \) are the points \( X \) and \( Z \).
   - Analogically, if we construct the circle \( t(H, \sqrt{HI \cdot HF}) \) and intersect it with the line \( VF \equiv YW \), we would get points \( Y \) and \( W \).
   - Now we can construct \( (XYZW) \) because we have constructed the points \( X, Y, Z, W \) and have \( I \).

6. **Constructing points \( A, B, C, D \):**
   - We can construct points \( A, B, C, D \) as the intersection points of the tangents to \( (XYZW) \) at \( X, Y, Z, W \).
   - In more detail, if \( l_x, l_y, l_z, l_w \) are the tangents to \( (XYZW) \) at \( X, Y, Z, W \) respectively, then:
     \[
     A = l_x \cap l_w, \quad B = l_x \cap l_y, \quad C = l_y \cap l_z, \quad D = l_z \cap l_w
     \]
   - One might say that in constructing \( X, Y, Z, W \) we can't distinguish \( X \) from \( Z \) and \( Y \) from \( W \), but this is irrelevant, because if we were to change \( l_x \) and \( l_z \) (for example), the intersection points mentioned above won't change, that is, \( A \) would be \( l_z \cap l_w \) in this case for example, but we would have constructed the 4 vertices of \( ABCD \), so we've restored it.