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

Given: A circle with centre O and diameter AB. Let PQ and RS be the tangents drawn at ends A and B respectively.

To prove: PQ ∥ RS

Proof:
Since OA is the radius and PQ is the tangent at A,
$$\angle OAP = 90° \quad \text{(Tangent ⊥ radius at point of contact)}$$

Since OB is the radius and RS is the tangent at B,
$$\angle OBS = 90°$$

Since ∠OAP = ∠OBS = 90°, and these are alternate interior angles (or co-interior angles each = 90°) formed by the transversal AB with lines PQ and RS,

$$\therefore PQ \parallel RS$$

Source: Chapter 10, Section 10.2 (Theorem 10.1)

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• The key theorem used is: tangent ⊥ radius at point of contact.
• Since both tangents are perpendicular to the same line (diameter AB), they are parallel to each other — this is the core logic.
• State "Given," "To Prove," and "Proof" clearly for full marks.
• Examiners award 1 mark for correctly applying the theorem at both ends, and 1 mark for concluding parallelism.