Question

Prove that If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle (i.e. they are concyclic).

Answer 1

Given: AB is a line segment and C and D are two points lying on the same side of AB such that ∠ACB = ∠ADB.

To prove: A, B, C and D are concyclic.

Proof:

If possible, suppose D does not lie on this circle. Let D´ be the point which lies on the circle.

Since, C and D´ are two point on the circle lying on the same side of AB.

∴ ∠ACB = ∠AD´B (Angles in the same segment are equal)
∠ADB = ∠AD´B (∠ACB = ∠ADB)

∴ An exterior of ΔDBD´ is equal to the interior opposite angle. But, an exterior angle of a triangle can never be equal to its interior opposite angle.

∴ ∠ADB = ∠AD´B

⇒ D coincides with D´.

⇒ D lies on the circle passing through the points A, B and C.

Hence, the points A, B, C and D are concyclic.