# 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.

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°.

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