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

Given: A chord AB of a circle with centre O. Tangents PQ and RS are drawn at points A and B respectively.

To Prove: ∠PAB = ∠RBA (tangents make equal angles with the chord)

Proof:

Let the tangents at A and B meet at an external point T.

Since tangents from an external point are equal (Theorem 10.2):
$$TA = TB$$

∴ Triangle TAB is isosceles.

$$\Rightarrow \angle TAB = \angle TBA \quad \text{...(1)}$$

Since PA is tangent at A:
$$\angle PAB = \angle TAB \quad \text{(same angle)} \quad \text{...(2)}$$

Since RB is tangent at B:
$$\angle RBA = \angle TBA \quad \text{(same angle)} \quad \text{...(3)}$$

From (1), (2), and (3):
$$\boxed{\angle PAB = \angle RBA}$$

Hence proved. The tangents drawn at the endpoints of a chord make equal angles with the chord.

Source: Chapter 10, Section 10.3 (Theorem 10.2)

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• The key idea is that the two tangents meet at an external point T, making TA = TB (Theorem 10.2), so △TAB is isosceles → base angles are equal.
• Examiners expect you to explicitly state "isosceles triangle" and cite equal tangents as the reason.
• Label the diagram logic clearly: ∠PAB = ∠TAB and ∠RBA = ∠TBA are the same angles, just named differently — don't skip this step.