Question

prove that the opposite sides of the quadrilateral acircumcribing a circle subtend supplementary angles at the centre

Answer 1

Answer:

Step-by-step explanation:

Let PQRS be a quadrilateral circumscribing a circle with centre O.

Join OA, OB, OC and OD.

Since the two tangents drawn from an external point to a circle subtend equal angles at the centre.

∴ ∠1 = ∠8

Similarly, ∠2 = ∠3, ∠4 = ∠5 and ∠6 = ∠7 ........ (1)

Now,

Sum of the angles at the centre = 360°

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

⇒ 2(∠1 + ∠2 + ∠5 + ∠6) = 360° [From equation (1)]

and 2(∠7 + ∠8 + ∠3 + ∠4) = 360°

⇒ ∠1 + ∠2 + ∠5 + ∠6 = 180°

and ∠3 + ∠4 + ∠7 + ∠8) = 180°

Thus, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles of 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

now,

∠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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