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 the circle. These four points are called _______

"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 the circle. These four points are called concyclic points.

Prove that any two vertices of a regular pentagon are concyclic. Click 'Yes' after proving this.

Here, △QRS ≅ △TSR (By SAS criteria) Therefore ∠RQS = ∠STR That is RS subtends the same angle at Q and T. This proves that any two vertices of regular pentagon are concyclic.

PQRS is a parallelogram. A circle going through P and Q is drawn such that it intersects PS at A and QR at B. Prove that A, B, R and S are concyclic points. Click 'Yes' after proving it.

PQRS is a parallelogram. A circle going through P and Q is drawn such that it intersects PS at A and QR at B. ∴Exterior ∠1 = Interior opposite ∠Q (Exterior angle property of cyclic quadrilateral) ∵ PQ ∥ RS and QR cuts them ∴∠Q + ∠R = 180⁰ (Sum of interior angles on both sides of transversal is 180⁰) or ∠1 + ∠R = 180⁰ [∵ ∠1 = ∠Q] ∴ ASRB is a cyclic quadrilateral. Hence, points A, B, R and S are concyclic points.

A, B and C are mid-points of the sides YZ, XZ and XY of a triangle and XD is perpendicular from X on YZ. Prove that A, B, C and D are concyclic. Click 'Yes' after proving it.

Join CD, BD, AC and AB. C and A are mid-points of XY and YZ. ∴ AC ∥ XZ Similarly AB ∥ XY ∴ XCAB is a parallelogram. ∴ ∠CXB = ∠CAB (opposite angles of the parallelogram are equal)......(1) XDY is a right angle triangle and DC is the median. CX = DC and ∠1 = ∠2.......(2) Similarly, ∠3 = ∠4.......(3) Adding (2) and (3) ∠1 + ∠3 = ∠2 + ∠4 ∠CDX + ∠XDB = ∠CXD + ∠BXD ∠CDB = ∠CXB ∠CDB = ∠CXB = ∠CAB (proved) Hence, A, B, C and D are concyclic points.

How many circles we can draw through four concyclic points?

We can draw only one circle passing through concyclic points.