Two Parallel Tangents of a Circle Meet a Third Tangent

Here we will prove that two parallel tangents of a circle meet a third tangent at points A and B. Prove that AB subtends a right angle at the centre.

Two Parallel Tangents of a Circle Meet a Third Tangent

Solution:

Given: CA, AB and EB are tangents to a circle with centre O. CA ∥ EB.

To prove: ∠AOB = 90°.

Proof:

Statement

Reason

1. AO bisects ∠CAD

⟹ ∠OAD = \(\frac{1}{2}\)∠CAD

1. The line joining the centre of a circle to the point of intersection of two tangents bisects the angle between the tangents.

2. BO bisects ∠DBE

⟹ ∠OBD = \(\frac{1}{2}\)∠DBE.

2. As in statement 1.

3. ∠CAD + ∠DBE = 180°

⟹ \(\frac{1}{2}\)∠CAD + \(\frac{1}{2}\)∠DBE = \(\frac{1}{2}\)180°

⟹ ∠OAD + ∠OBD = 90°.

3. Co. interior angles and CA ∥ EB.



Using statements 1 and 2 in statement 3.

4. Therefore, ∠AOB = 180° - (∠OAD + ∠OBD)

                            = 180° - 90°

                            = 90°. (proved).

4. Sum of three angles of a triangle is 180°.





10th Grade Math

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