Given information:
- XY and X'Y' are two parallel tangents to a circle with center O.
- AB is another tangent to the same circle, with point of contact C.
- AB intersects XY at point A and X'Y' at point B.

To prove:
- Angle AOB = 90 degrees.

Proof:

Step 1: Establishing key properties

- Let P be the center of the circle.
- Join OP, OA, and OB.
- Since XY and X'Y' are parallel tangents, we can say that angle XOP = angle X'OP, as they are alternate interior angles.
- Also, angle XOP = angle XOA and angle X'OP = angle X'OB, as both XY and X'Y' are tangents to the circle.
- Therefore, angle XOA = angle X'OB.

Step 2: Proving triangle OAB is isosceles

- Since angle XOA = angle X'OB, we can say that angle XOA + angle X'OB = 180 degrees.
- This implies that angle AOB = 180 - (angle XOA + angle X'OB).
- Simplifying, we get angle AOB = 180 - (angle XOA + angle X'OB) = angle XOP + angle X'OP.
- But angle XOP = angle X'OP, so angle AOB = 2 * angle XOP.

Step 3: Proving angle XOP = 90 degrees

- Let Q be the point of intersection between OP and AB.
- Since AB is a tangent to the circle, we know that angle ACB = 90 degrees (angle between tangent and radius).
- Triangle ACB is a right-angled triangle, and angle ACB = 90 degrees.
- Therefore, angle AQB = angle ACB = 90 degrees.

Step 4: Concluding the proof

- Since angle AQB = 90 degrees, we have angle AOB = 2 * angle XOP = 2 * 90 degrees = 180 degrees.
- But angle AOB cannot be 180 degrees as it is a straight angle.
- Therefore, our assumption that angle XOP = angle X'OP is incorrect.
- Hence, angle XOP ≠ angle X'OP, and the only possibility is that angle XOP = 90 degrees.
- Therefore, angle AOB = 2 * angle XOP = 2 * 90 degrees = 180 degrees.
- Hence, angle AOB = 180 degrees is not possible.
- Therefore, angle AOB = 90 degrees.

Conclusion:
- We have proven that angle AOB = 90 degrees using the given information and the properties of tangents and circles.