Name: Tangents From an Exterior Point: Illustration
Uploaded: 2019-07-19
Duration: 4 min 47 s
Description: If you lift a circular object with a pair of tongs, then the tongs are two tangents to the circular object. Let us study the properties of tangents drawn fro...

Question

Prove that 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 centre $O$ . Also, the sides AB, BC, CD and DA touches the circle at point $P, Q, R,$ and $S$ respectively.

Now join the vertices of the quadrilateral $A B C D$ to the center O of the circle.

In triangles $∆ O A P$ and $∆ O A S$

$A P = A S$ (length of tangents from same point)

$O P = O S$ (radius of the same circle)

$O A = OA$ (common side)

Hence $∆$ OAP $≅ ∆$ OAS (by SSS congruency)

$\Rightarrow ∠ P O A = ∠ SOA$ (by CPCT)

$\Rightarrow ∠ 1 = ∠ 8$ …………………(i)

Similarly, we can prove

$∠ 2 = ∠ 3$ ……………………(ii)

$∠ 4 = ∠ 5$ ……………………(iii) and

$∠ 6 = ∠ 7$ ……………………(iv)

On adding (i), (ii), (iii) and (iv), we get

$∠ 1 + ∠ 2 + ∠ 3 + ∠ 4 + ∠ 5 + ∠ 6 + ∠ 7 + ∠ 8 = 360^{o}$

$\Rightarrow (∠ 1 + ∠ 8) + (∠ 2 + ∠ 3) + (∠ 4 + ∠ 5) + (∠ 6 + ∠ 7) = 360^{o}$

$\Rightarrow 2 ∠ 1 + 2 ∠ 2 + 2 ∠ 5 + 2 ∠ 6 = 360^{o}$

$\Rightarrow 2 (∠ 1 + ∠ 2) + 2 (∠ 5 + ∠ 6) = 360^{o}$

$\Rightarrow (∠ 1 + ∠ 2) + (∠ 5 + ∠ 6) = 180^{o}$

$\Rightarrow ∠ A O B + ∠ C O D = 180^{o}$

Similarly, $∠ B O C + ∠ D O A = 180^{o}$

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

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

Important Questions on Circles

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