prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle
Answers
Given : Let ABCD be the quadrilateral circumscribing the circle with centre O. ABCD touches the circle at points P,Q,R and S
To prove: Opposite sides subtend supplementary angles at centre i.e. ∠ AOB + ∠ COD = 180° & ∠ AOD + ∠ BOC = 180°
Construction: Join OP, OQ, OR & OS
Proof: Let us rename the angles In Δ AOP and Δ AOS AP = AS AO = AO OP = OS ∴ Δ AOP ≅∆ AOS ∠ AOP = ∠ AOS i.e. ∠ 1 = ∠ 8
Similarly, we can prove ∠2 = ∠3 ∠5 = ∠4 ∠6 = ∠7 Now ∠ 1 + ∠ 2 + ∠ 3 + ∠ 4 + ∠ 5 + ∠ 6 + ∠ 7 + ∠ 8 = 360° ∠ 1 + ∠ 2 + ∠ 2 + ∠ 5 + ∠ 5 + ∠ 6 + ∠ 6 + ∠ 1 = 360° 2 (∠ 1 + ∠ 2 + ∠ 5 + ∠ 6) = 360° ∠ 1 + ∠ 2 + ∠ 5 + ∠ 6 = (360°)/2 (∠ 1 + ∠ 2) + (∠ 5 + ∠ 6) = 180° ∠ AOB + ∠ COD =180°
Hence both angle are supplementary Similarly, we can prove ∠ BOC + ∠ AOD =180° Hence proved
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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