Proof:
Given:
- XY and X'Y' are two parallel tangents to a circle with center O.
- AB is another tangent with the point of contact C.
- A and B are the points where AB intersects XY and X'Y' respectively.

To Prove:
- ∠AOB = 90°

Proof:

1. Let's draw a diagram to visualize the given information.

2. Construction:
- Join OA, OB, OC, OX, and OY.

3. As XY and X'Y' are tangents to the circle at points A and B respectively, the radii OA and OB will be perpendicular to the tangents at their respective points of contact (property of tangents).

4. Claim: ∠OAC = ∠OBC = 90°
Proof of Claim:
- In triangle OAC, OA and OC are radii of the circle, so they are equal in length (property of circle).
- Similarly, in triangle OBC, OB and OC are radii of the circle, so they are equal in length.
- Therefore, triangles OAC and OBC are congruent by the side-side-side congruence criterion.
- Since AC and BC are corresponding sides of congruent triangles, ∠OAC = ∠OBC (property of congruent triangles).
- Also, since AC and BC are perpendicular to the tangents XY and X'Y' respectively, they are perpendicular to the radii OA and OB at their respective points of contact.
- Hence, ∠OAC = ∠OBC = 90°.

5. Since ∠OAC = ∠OBC = 90°, the quadrilateral AOCB is a cyclic quadrilateral with OC as its diameter (property of a cyclic quadrilateral).
- Therefore, ∠AOB = 180° - ∠ACB (property of a cyclic quadrilateral).

6. In triangle ACB, AC is a tangent to the circle at point A and BC is a tangent to the circle at point B.
- The angle between a tangent and a chord drawn from the point of contact is always 90° (property of tangents).
- Therefore, ∠ACB = 90°.

7. Substituting the value of ∠ACB in ∠AOB = 180° - ∠ACB, we get ∠AOB = 180° - 90° = 90°.

8. Hence, we have proved that ∠AOB = 90°.

This completes the proof.