The Perpendicular Bisectors and the Circumcenter of a Triangle

The perpendicular bisectors of the sides of a triangle are concurrent at a point called circumcenter.

This point of concurrency is the center of the circle circumscribing the triangle and is equidistant from the vertices of the triangle.

This post is the second to the last part of a series of posts about the points of concurrency of a triangle's parts.

Here's a list of my previous posts about concurrent parts of a triangle:

• orthocenter: the point of concurrency of a triangle's altitudes (see Altitudes and Orthocenter of a Triangle)
• centroid: the point of concurrency of a triangle's medians (see Medians And Centroid Of A Triangle)
• incenter: the point of concurrency of a triangle's angle bisectors (see The Angle Bisectors and the Incenter of a Triangle)

The Problem

Given the vertices A(-5, 2), B(2, 3) and C(-1, -6) of triangle ABC, find the point of concurrency of the perpendicular bisectors of its sides. Find the equation of the circle circumscribed about the triangle.

Discussion

Referring to the given illustration above, line segments OP, OQ and OR are the perpendicular bisectors of the sides of scalene triangle ABC.

Their point of concurrency, point O, is called circumcenter because it is the center of the circle circumscribing the triangle.

Line segments AO, BO and CO are the radii of circle O.

The problem illustrates the fact that any triangle can always be circumscribed by a circle.

In other words, a set of three noncollinear points which are on the same circle (concyclic) can always be found.

Solution

Coordinates of Circumcenter: (-1, -1)



Midpoints of the Sides

• side AB: P(- 3/2, 5/2)
• side BC: Q(1/2, - 3/2)
• side AC: R(-3, -2)

Distance Between Two Points

• distance_AB = √ 50
• distance_BC = √ 90
• distance_AC = √ 80
• distance_AO = distance_BO = distance_CO = √ 25
• distance_AP = distance_BP = √ 25/2
• distance_BQ = distance_CQ = √ 45/2
• distance_AR = distance_CR = √ 20

Equations of the Line Segments

• side AB: x - 7y + 19 = 0; x-intercept = -19; y-intercept = 19/7
• side BC: 3x - y - 3 = 0; x-intercept = 1; y-intercept = -3
• side AC: 2x + y + 8 = 0; x-intercept = - 4; y-intercept = - 8
• perpendicular bisector OP: 7x + y + 8 = 0; x-intercept = - 8/7; y-intercept = - 8
• perpendicular bisector OQ: x + 3y + 4 = 0; x-intercept = - 4; y-intercept = - 4/3
• perpendicular bisector OR: x - 2y - 1 = 0; x-intercept = 1; y-intercept = - 1/2
• radius AO: 3x + 4y + 7 = 0; x-intercept = - 7/3; y-intercept = - 7/4
• radius BO: 4x - 3y + 1 = 0; x-intercept = - 1/4; y-intercept = 1/3
• radius CO: x = - 1; x-intercept = - 1

Equation of the Circle:

( x + 1 )² + ( y + 1 )² = (√ 25)²

x² + y² + 2x + 2y - 23 = 0

Slopes of the Line Segments

• m_AB = 1/7
• m_BC = 3
• m_AC = - 2
• m_AO = - 3/4
• m_BO = 4/3
• m_OP = - 7
• m_OQ = - 1/3
• m_OR = 1/2

Interior Angles of the Triangle

• angle A = 71.6°
• angle B = 63.4°
• angle C = 45°

The Angle Bisectors and the Incenter of a Triangle

The problem presented here serves as an illustration of the properties of the following:

• angle bisectors of a triangle
• lines tangent to a circle
• a circle inscribed in a triangle

The Problem

Given the vertices A(0, 2), B(3, 5) and C(4, -2) of right triangle ABC, show analytically that the bisectors of this triangle's angles are concurrent at a point. Show also that this point of concurrency is the center of the inscribed circle.

Discussion

In the illustration above, line segments AO, BO and CO are the angle bisectors of right triangle ABC.

Their point of concurrency, point O, is called incenter because it is the center of the inscribed circle, circle O.

Circle O is tangent to sides AB, BC and AC at points D, E and F, respectively.

Since line segments DO, EO and FO are the radii of circle O and are perpendicular to the three sides of the triangle, point O is equidistant from the three sides of triangle ABC.

As for the tangent lines, there is a theorem that states: from a point outside the circle, lines tangent to the circle are congruent.

Hence,

• line segment AD = line segment AF
• line segment BD = line segment BE
• line segment CE = line segment CF

Solution

Coordinates of Incenter: (2, 2)

Distance Between Two Points

• distance_AB = √ 18
• distance_BC = √ 50
• distance_AC = √ 32
• distance_DO = distance_EO = distance_FO = √ 2
• distance_AD = distance_AF = √ 2
• distance_BD = distance_BE = √ 8
• distance_CE = distance_CF = √ 18

Equations of the Line Segments

• side AB: x - y + 2 = 0; x-intercept = - 2; y-intercept = 2
• side BC: 7x + y - 26 = 0; x-intercept = 26/7; y-intercept = 26
• side AC: x + y - 2 = 0; x-intercept = 2; y-intercept = 2
• angle bisector AO: y - 2 = 0; x-intercept = 2
• angle bisector BO: 3x - y - 4 = 0; x-intercept = 4/3; y-intercept = - 4
• angle bisector CO: 2x + y - 6 = 0; x-intercept = 3; y-intercept = 6
• radius DO: x + y - 4 = 0; x-intercept = 4; y-intercept = 4
• radius EO: x - 7y + 12 = 0; x-intercept = - 12; y-intercept = 12/7
• radius FO: x - y = 0; x-intercept = 0; y-intercept = 0

Equation of the Circle:

( x - 2 )² + ( y - 2 )² = (√ 2)²

x² + y² - 4x - 4y + 6 = 0

Slopes of the Line Segments

• m_AB = 1
• m_BC = - 7
• m_AC = - 1
• m_AO = 0
• m_BO = 3
• m_CO = - 2
• m_DO = - 1
• m_EO = 1/7
• m_FO = 1

Interior Angles of the Triangle

• angle A = 90.0°
• angle B = 53.13°
• angle C = 36.87°
• angle BAO = angle CAO = 45.0°
• angle ABO = angle CBO = 26.57°
• angle BCO = angle ACO = 18.43°