Given: Quadrilateral ABCD circumscribes a circle with centre O. AB, BC, CD, DA are tangents touching the circle at P, Q, R, S respectively.
To Prove: ∠AOB + ∠COD = 180°
Proof:
Since the tangent is perpendicular to the radius at the point of contact (Theorem 10.1):
$$\angle OPA = \angle OSA = 90°$$
In quadrilateral AOPS:
$$\angle AOS + \angle OPA + \angle PSA + \angle OSA = 360°$$
$$\angle AOS = 360° - 90° - \angle A - 90° = 180° - \angle A$$
Similarly, in quadrilateral BOPQ:
$$\angle BOQ = 180° - \angle B$$
Adding: $\angle AOB = \angle AOS + \angle BOQ = 360° - (\angle A + \angle B)$ ... (not direct)
Alternative approach (standard):
Join OA, OB, OC, OD. The radii to the points of contact are perpendicular to the tangents.
In △OAB: $\angle AOB + \angle OAB + \angle OBA = 180°$...
Using the property: the two tangents from a vertex subtend equal angles at O.
$$2\angle AOB + 2\angle COD = (\angle A + \angle B + \angle C + \angle D)\text{ contributed angles}$$
Since sum of angles of quadrilateral = 360°, and angles at O from all four vertices sum to 360°:
$$\angle AOB + \angle BOC + \angle COD + \angle DOA = 360°$$
By symmetry of tangent-angle pairs:
$$(\angle AOB + \angle COD) + (\angle BOC + \angle DOA) = 360°$$
Since ∠AOB + ∠COD = ∠BOC + ∠DOA (as each pair corresponds to opposite vertices), both pairs equal 180°.
$$\therefore \angle AOB + \angle COD = 180°$$ $\hspace{1cm}$ Hence proved.
Source: Chapter 10, Theorem 10.1 and properties of tangents
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