Question

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

Accepted Answer

Let $A B C D$ be a quadrilateral circumscribing a circle with $O$ such that it touches the circle at point $P, Q, R, S$ .

Join the vertices of the quadrilateral $A B C D$ to the center of the circle.

In $Δ O A P$ and $Δ O A S$ ,

$A P = A S$ (Tangents from the same point)

$O P = O S$ (Radii of the circle)

$O A = O A$ (Common side)

$Δ O A P ≅ Δ O A S$ ( $S S S$ congruence condition)

$∴ ∠ P O A = ∠ A O S \Rightarrow ∠ 1 = ∠ 8$

Similarly we get, $∠ 2 = ∠ 3, ∠ 4 = ∠ 5 and ∠ 6 = ∠ 7$ .

Adding all these angles,

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

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

$\Rightarrow 2 ∠ 1 + 2 ∠ 2 + 2 ∠ 5 + 2 ∠ 6 = 360 °$

$\Rightarrow 2 (∠ 1 + ∠ 2) + 2 (∠ 5 + ∠ 6) = 360 °$

$\Rightarrow (∠ 1 + ∠ 2) + (∠ 5 + ∠ 6) = 180 °$

$\Rightarrow ∠ A O B + ∠ C O D = 180 °$

Similarly, we can prove that $∠ B O C + ∠ D O A = 180 °$ .

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

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

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