sol it--------- on copy pge 13th
Answers
To Prove : ∠AOB + ∠COD = 180°;∠BOC + ∠DOA = 180°
Given: ABCD is circumscribing the circle.
Proof:
Let ABCD be a quadrilateral circumscribing a circle centred at O such that it touches the circle at point P, Q, R, S.
join the vertices of the quadrilateral ABCD to the centre of the circle.
Consider 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)
And thus, ∠ POA = ∠ AOS
- ∠1 = ∠ 8 Similarly,
- ∠2 = ∠ 3
- ∠4 = ∠ 5
- ∠6 = ∠ 7
∠1 + ∠ 2 + ∠ 3 + ∠ 4 + ∠ 5 + ∠ 6 + ∠ 7 + ∠ 8 = 360°
(∠ 1 + ∠ 8) + (∠ 2 + ∠ 3) + (∠ 4 + ∠ 5) + (∠ 6 + ∠ 7) = 360°
2∠ 1 + 2∠ 2 + 2∠ 5 + 2∠ 6 = 360°
2(∠ 1 + ∠ 2) + 2(∠ 5 + ∠ 6) = 360°
(∠ 1 + ∠ 2) + (∠ 5 + ∠ 6) = 180°
∠ AOB + ∠ COD = 180°
Similarly, we can prove that ∠ BOC + ∠ DOA = 180°
Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
[Proved] Q.E.D
Answer:
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