Question

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

Answer 1

Let ABCD be a quadrilateral circumscribing a circle with centre O. 
Now join AO, BO, CO, DO.
From the figure, ∠DAO = ∠BAO [Since, AB and AD are tangents] 
Let ∠DAO = ∠BAO = 1 
Also ∠ABO = ∠CBO [Since, BA and BC are tangents] 
Let ∠ABO = ∠CBO = 2 
Similarly we take the same way for vertices C and D
Sum of the angles at the centre is 360° 
Recall that sum of the angles in quad. ABCD = 360° 
⇒ 2(1 + 2 + 3 + 4) = 360° 
⇒ 1 + 2 + 3 + 4 = 180° 
In ΔAOB, ∠BOA = 180 – (a + b)
In ΔCOD, ∠COD = 180 – (c + d)
Angle BOA + angle COD = 360 – (a + b + c + d) 
= 360°  – 180° 
= 180° 
Hence AB and CD subtend supplementary angles at O
Thus, 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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