Given: XY and X'Y' are parallel tangents to a circle with centre O, touching at P and Q respectively. Tangent AB touches the circle at C, meeting XY at A and X'Y' at B.
To Prove: ∠AOB = 90°
Proof:
Since AP and AC are tangents from external point A to the circle:
$$AP = AC \quad \text{(tangents from same external point are equal)}$$
∴ OA bisects ∠PAC (centre lies on angle bisector of tangents from an external point)
$$\Rightarrow \angle OAP + \angle OAC = \angle PAC$$
$$\Rightarrow \angle OAC = \frac{1}{2}\angle PAC \quad \ldots(1)$$
Similarly, from external point B: BQ = BC
∴ OB bisects ∠QBC
$$\Rightarrow \angle OBC = \frac{1}{2}\angle QBC \quad \ldots(2)$$
Since XY ∥ X'Y', PA ∥ QB (both on parallel lines), so:
$$\angle PAC + \angle QBC = 180° \quad \text{(co-interior angles)}$$
$$\Rightarrow \frac{1}{2}\angle PAC + \frac{1}{2}\angle QBC = 90°$$
$$\Rightarrow \angle OAC + \angle OBC = 90° \quad \ldots(3)$$
In △AOB:
$$\angle AOB + \angle OAB + \angle OBA = 180°$$
$$\angle AOB + \angle OAC + \angle OBC = 180°$$
$$\angle AOB = 180° - 90° = \mathbf{90°} \quad \blacksquare$$
Source: Chapter 10, Section 10.3 (Exercise 10.2, Q.9)
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