Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Question

Philoid · Accepted Answer

Let ABCD be a quadrilateral circumscribing a circle centred at O such that it touches the circle at point P, Q, R, S.

join the vertices of the quadrilateral ABCD to the centre of the circle.

Consider ΔOAP and ΔOAS,

AP = AS (Tangents from the same point)

OP = OS (Radii of the same circle)

OA = OA (Common side)

ΔOAP ≅ ΔOAS (SSS congruence criterion)

And thus, ∠ POA = ∠ AOS

∠1 = ∠ 8 Similarly,

∠2 = ∠ 3

∠4 = ∠ 5

∠6 = ∠ 7

∠1 + ∠ 2 + ∠ 3 + ∠ 4 + ∠ 5 + ∠ 6 + ∠ 7 + ∠ 8 = 360°

(∠ 1 + ∠ 8) + (∠ 2 + ∠ 3) + (∠ 4 + ∠ 5) + (∠ 6 + ∠ 7) = 360°

2∠ 1 + 2∠ 2 + 2∠ 5 + 2∠ 6 = 360°

2(∠ 1 + ∠ 2) + 2(∠ 5 + ∠ 6) = 360°

(∠ 1 + ∠ 2) + (∠ 5 + ∠ 6) = 180°

∠ AOB + ∠ COD = 180°

Similarly, we can prove that ∠ BOC + ∠ DOA = 180°

Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.