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.