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Theorem : If a pair of opposite angles  a quadrilateral is supplementary, then the quadrilateral is cyclic. 

Given: In ABCD, ∠A + ∠C = 180° 

To prove: ABCD is a cyclic quadrilateral.

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Proof: 

(Indirect method) 

Suppose ABCD is not a cyclic quadrilateral. 

We can still draw a circle passing through three non-collinear points A, B,

. Case I: Point C lies outside the circle. 

Then, take point E on the circle such that D – E – C.

∴ ⟂ABED is a cyclic quadrilateral. 

∠DAB + ∠DEB = 180° (i) [Opposite angles of a cyclic quadrilateral are supplementary] 

∠DAB + ∠DCB = 180° (ii) [Given] 

∴ ∠DAB + ∠DEB = ∠DAB + ∠DCB [From (i) and (ii)] 

∴ ∠DEB = ∠DC

But, ∠DEB ≠ ∠DCB as ∠DEB is the exterior angle of ∆BEC. 

∴ Our supposition is wrong. 

∴ ABCD is a cyclic quadrilateral. 

Case II: Point C lies inside the circle. Then, take point E on the circle such that D – C – E

∴ □ABED is a cyclic quadrilateral. 

∠DAB + ∠DEB = 180° (iii) [Opposite angles of a cyclic quadrilateral are supplementary] 

∠DAB + ∠DCB = 180° (iv) [Given] 

∴ ∠DAB + ∠DEB = ∠DAB + ∠DCB [From (iii) and (iv)] 

∴ ∠DEB = ∠DCB 

But ∠DEB ≠ ∠DCB as ∠DCB is the exterior angle of ∆BCE. 

∴ Our supposition is wrong. 

∴ □ABCD is a cyclic quadrilateral.

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