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°