# If a pair of opposite angles a quadrilateral is supplementary, then the quadrilateral is cyclic. Given: In ABCD, ∠A + ∠C = 180°

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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.