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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