**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.