4. In the plane containing the segment $B D$, find the set of all vertices $A$ of convex quadrilaterals $A B C D$ for which the following conditions are simultaneously satisfied:

a) the center $O_{C}$ of the incircle of triangle $B C D$ lies on the circumcircle of triangle $A B D$,

b) the center $O_{A}$ of the incircle of triangle $A B D$ lies on the circumcircle of triangle $B C D$.
Solution: Let $\alpha=|\angle B A D|$ and $\gamma=|\angle B C D|$. We have

$$
\left|\angle B O_{C} D\right|=180^{\circ}-\left(\left|\angle O_{C} B D\right|+\left|\angle O_{C} D B\right|\right)=
$$

$$
=180^{\circ}-\frac{1}{2}(|\angle C B D|+|\angle C D B|)=90^{\circ}+\frac{1}{2}
$$

Similarly, $\left|\angle B O_{A} D\right|=90^{\circ}+\frac{1}{2} \alpha$. Since points $A$ and $O_{C}$ lie in opposite half-planes with the boundary line $B 1$, condition a) of the problem is equivalent to the sum of the magnitudes of angles $B A D$ and $B O_{C} D$ being $180^{\circ}$, i.e., $\alpha+\left(90^{\circ}+\frac{1}{2} \gamma\right)=180^{\circ}$. Similarly, we conclude that condition b) is satisfied if and only if $\gamma+\left(90^{\circ}+\frac{1}{2} \alpha\right)=180^{\circ}$. The found pair of equations has a unique solution $\alpha=\gamma=60^{\circ}$. Therefore, points $A$ and $C$ lie on two different circular arcs from which the segment $B D$ is seen at an angle of $60^{\circ}$. On the other hand, if we choose any interior point $A$ of one of these arcs, we can choose a point $C$ on the other arc such that $A B C D$ is a convex quadrilateral (it suffices to complete the triangle $A B D$ to a parallelogram $A B C D$).

Answer: The set of vertices $A$ consists of the interior points of two circular arcs from which the segment $B D$ is seen at an angle of $60^{\circ}$.

Grading instructions: If the solver does not explain why every point $A$ in the found set is a vertex of some suitable quadrilateral $A B C D$, award no more than 5 points.