The given points are:
- A(5, 5)
- B(6, 4)
- C(2, 4)
- D(7, 1)


To prove that these points are concyclic, we need to show that the perpendicular bisectors of the line segments AB, BC, CD, and DA intersect at a common point.


The midpoint of AB is M, which can be found by taking the average of the x-coordinates and the average of the y-coordinates of A and B:
- M = ((5+6)/2, (5+4)/2) = (5.5, 4.5)

The slope of AB is:
- m_AB = (4-5)/(6-5) = -1

The negative reciprocal of the slope is the slope of the perpendicular bisector:
- m_perpendicular_AB = 1

Using the point-slope form of a line, the equation of the perpendicular bisector of AB is:
- y - 4.5 = 1(x - 5.5)
- y - 4.5 = x - 5.5
- y = x - 1


The midpoint of BC is N, which can be found by taking the average of the x-coordinates and the average of the y-coordinates of B and C:
- N = ((6+2)/2, (4+4)/2) = (4, 4)

The slope of BC is:
- m_BC = (4-4)/(2-6) = 0

The negative reciprocal of the slope is the slope of the perpendicular bisector:
- m_perpendicular_BC = undefined

Since the slope is undefined, the perpendicular bisector of BC is a vertical line passing through the midpoint N:
- x = 4


The midpoint of CD is O, which can be found by taking the average of the x-coordinates and the average of the y-coordinates of C and D:
- O = ((2+7)/2, (4+1)/2) = (4.5, 2.5)

The slope of CD is:
- m_CD = (1-4)/(7-2) = -3/5

The negative reciprocal of the slope is the slope of the perpendicular bisector:
- m_perpendicular_CD = 5/3

Using the point-slope form of a line, the equation of the perpendicular bisector of CD is:
- y - 2.5 = (5/3)(x - 4.5)
- 3y - 7.5 = 5x - 22.5
- 5x - 3y = 15


The midpoint of DA is P, which can be found by taking the average of the x-coordinates and the average of the y-coordinates of D and A:
- P = ((7+5)/2, (1+5)/2) = (6, 3)

The slope of DA is:
- m_DA = (3-