Question

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

Answer 1

Given: AB is a line segment, which subtends equal angles at two points C and D. i.e., ∠ACB = ∠ADB. 

To Prove: The points A, B, C and D lie on a circle. 

Proof: Let us draw a circle through the points A, C and B. 

Suppose it does not pass through the point D. 

Then it will intersect AD (or extended AD) at a point, say E (or E’). 

If points A,C,E and B lie on a circle, 

∠ACD = ∠AEB [∴ Angles in the same segment of circle are equal] 

But it is given that ∠ACB = ∠ADB 

Therefore, ∠AEB = ∠ADB 

This is possible only when E coincides with D. [As otherwise ∠AEB > ∠ADB] 

Similarly, E’ should also coincide with D. So A, B, C and D are concyclic