Mathematics — Standard (041) — CBSE Class 10 board question

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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• The key theorem used is: tangents from an external point are equal in length, and OA bisects the angle between the two tangents from A (stated in Remark 2 after Theorem 10.2).
• The co-interior angles property (∠PAC + ∠QBC = 180°) comes from XY ∥ X'Y' with AB as transversal — this is the critical step many students miss.
• Examiners award marks for: stating equal tangents, using angle bisector property, applying co-interior angles, and concluding with angle sum of triangle.