Given the rectangle $A B C D$. Draw the diagonals and the angle bisector from vertex $C$. The angle bisector intersects the side $A B$ at point $E$, the diagonal $B D$ at point $F$, and the intersection of the diagonals is point $O$.

What angle does the angle bisector form with the diagonal $A C$, given that $\Delta E O B$ and $\Delta E O F$ are both isosceles?
Due to the right angles and the angle bisectors, $B E = B C$. On the other hand, $B A > B E = B C$, because if $E$ were to coincide with $A$, then $F$ would coincide with $O$, and we could not speak of the triangle $E O F$. Thus, $O E < O A = O B$, so the axis of symmetry of the triangle $E O B$ does not pass through $O$, but either through $E$ or $B$.
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Considering the first possibility, $E O = E B = B C = 2 \cdot O G$ would be true, where $G$ is the midpoint of side $A B$. Furthermore, as the exterior angle of triangle $E O B$, $\angle O E A = 2 \cdot \angle O B A < 90^{\circ}$, because $\angle O B A < 90^{\circ}$, so $E$ would be on the segment $G B$. Thus, $\angle O E G = 30^{\circ}$, $\angle E O B = 15^{\circ}$, and $\angle O E F = 180^{\circ} - 45^{\circ} - 30^{\circ} = 105^{\circ}$, meaning that triangle $O E F$ would not be isosceles, as it does not have two equal angles.

However, if we try the possibility that the axis of symmetry of triangle $E O B$ passes through $B$, then $B C = B E = B O = C O$, so $\angle O C B = 60^{\circ}$ and the desired angle $\angle A C E = 15^{\circ}$. Then $\angle O E C = 30^{\circ}$, as it is an inscribed angle in the circle centered at $B$ passing through $E$, so $\angle O E B = 75^{\circ}$ and $\angle O B E = 30^{\circ}$, which means $\angle E O B$ is also $75^{\circ}$, so $E O = E F$, and triangle $E O F$ is isosceles, which meets the requirements of the problem. - We have completed the solution.