Two Parallel Tangents of a Circle Meet a Third Tangent

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°.