**Given:** Points B and C lie on the same side of the line AD ∠ABD = ∠ACD

**To prove:** Points A, B, C, D are concyclic.

**i.e., **□ABCD is a cyclic quadrilateral.

**Proof: **

Suppose points A, B, C, D are not concyclic points.

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

**Case I:** Point C lies outside the circle. Then, take point E on the circle such that A – E – C.

∠ABD ≅ ∠AED (i) [Angles inscribed in the same arc]

∠ABD ≅ ∠ACD (ii) [Given]

∴ ∠AED ≅ ∠ACD [From (i) and (ii)]

∴ ∠AED ≅ ∠ECD [A – E – C]

But, ∠AED ≅ ∠ECD as ∠AED is the exterior angle of ∆ECD.

∴ Our supposition is wrong.

∴ Points A, B, C, D are concyclic points.

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

∠ABD ≅ ∠AED (iii) [Angles inscribed in the same arc]

∠ABD ≅ ∠ACD (iv) [Given]

∴ ∠AED ≅ ∠ACD [From (iii) and (iv)]

∴ ∠CED ≅ ∠ACD [A – C – E]

But, ∠CED ≅ ∠ACD as ∠ACD is the exterior angle of ∆ECD.

∴ Our supposition is wrong.

∴ Points A, B, C, D are concyclic points.