To prove that the quadrilateral OABC is a square given the conditions of the problem, we can follow these steps:

1. Given that \( \angle xOy = 90^\circ \) and \( Ot \) is the angle bisector, we know that \( \angle xOt = \angle yOt \).

2. The point \( A \) is chosen on the line \( Ot \), and the lines \( AB \) and \( AC \) are drawn such that \( AB \perp Ox \) and \( AC \perp Oy \).

3. Since \( AB \perp Ox \) and \( AC \perp Oy \), it follows that both \( AB \) and \( AC \) are perpendicular to the x and y axes, respectively.

4. The coordinates of \( A\) can be represented as \( (a,0) \), where \( a \) is the x-coordinate. Therefore, the coordinates of \( B\) and \( C\) can be described as follows:
- Since \( AB\) is perpendicular to the x-axis, \( B\) will have coordinates \( (a, b) \).
- Since \( AC\) is perpendicular to the y-axis, \( C\) will have coordinates \( (c, 0) \).

5. The distance \( AB = AC = a \) implies that both segments are equal in length from point \( A \) to points \( B \) and \( C \).

6. Finally, to confirm that triangle \( OABC \) is a square, we need to check that \( OB \) is horizontal and \( OC \) is vertical (which is the case as established by how we've defined the perpendicular lines).

7. Since \( O \) is at the origin, we find:
- \( O \) at \( (0,0) \)
- \( B \) at \( (a, b) \)
- \( C \) at \( (c, 0) \)

8. Because both \( B \) and \( C \) have equal lengths from \( O \) and are perpendicular, the quadrilateral \( OABC \) must be a square, as all sides are equal and all angles are right angles.

Thus, we can conclude that the quadrilateral \( OABC \) is indeed a square, based on the above geometrical properties and conditions.