Question

prove that opposite sides of a quadrilateral circumscribing a circle subtended supplementary angles at the centre of the circle

Answer 1

Let ABCD be a quadrilateral circumscribing a circle centered at O such that it touches the circle at point P, Q, R, S. Let us join the vertices of the quadrilateral ABCD to the center of the circle.

In 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)

∠POA = ∠AOS  (CPCT)

or ∠1 = ∠8

Similarly,

∠2 = ∠3

∠4 = ∠5

∠6 = ∠7

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

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

2∠1 + 2∠2 + 2∠5 + 2∠6 = 360º

2(∠1 + ∠2) + 2(∠5 + ∠6) = 360º

(∠1 + ∠2) + (∠5 + ∠6) = 180º

∠AOB + ∠COD = 180º

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

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

Answer 2

Hey Mate :D

Your Answer :---

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

Let us join the vertices of the quadrilateral ABCD to the center of the circle.

[ plz see attached file also :) ]

Consider ∆OAP and ∆OAS,

AP = AS (Tangents from the same point)

OP = OS (Radii of the same circle)

OA = OA (Common side)

∆OAP ≅ ∆OAS (SSS congruence criterion)

Therefore, A ↔ A, P ↔ S, O ↔ O

And thus,

∠POA = ∠AOS ∠1 = ∠8 Similarly, ∠2 = ∠3 ∠4 = ∠5 ∠6 = ∠7

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

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

2 ∠1 + 2 ∠2 + 2∠ 5 + 2∠ 6 = 360º

2( ∠1 + ∠ 2) + 2( ∠5 + ∠ 6) = 360º

( ∠1 + ∠2 ) + ( ∠5 + ∠ 6) = 180º

∠AOB + ∠COD = 180º

Similarly, we can prove that

∠BOC + ∠DOA = 180º

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

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