Incenter of a triangle

1 Star2 Stars3 Stars4 Stars5 Stars (No Ratings Yet)
Loading...

Drawing of the incenter of a triangle as the intersection of the three bisectors

The incenter of a triangle (I) is the point where the three interior angle bisectors (Ba, Bb y Bc) intersect.

The angle bisector of a triangle is a line segment that bisects one of the vertex angles of a triangle, and it ends on the corresponding opposite side.

Drawing of the three bisectors of a triangle, the incenter and the inscribed circle

As we can see in the picture above, the incenter of a triangle (I) is the center of its inscribed circle (or incircle) which is the largest circle that will fit inside the triangle.

The radius (or inradius) of the incircle is found by the formula:

Formula for the inradius inscribed in the triangle with center at the incenter

Where is the Incenter of a Triangle Located?

The incenter (I) of a triangle is always inside it.

Download this calculator to get the results of the formulas on this page. Choose the initial data and enter it in the upper left box. For results, press ENTER.

Triangle-total.rar         or   Triangle-total.exe      

Note. Courtesy of the author: José María Pareja Marcano. Chemist. Seville, Spain.

Relationship between Inradius and Area

The inradius (or incircle’s radius) is related to the area of the triangle to which its circumference is inscribed by the relation:

Formula for the ratio of radius, area and semiperimeter

If is a right triangle this relation between inradius and area is:

Radius formula in the right triangle

Incenter Theorem

Drawing of the incenter theorem

The incenter I of a triangle Δ ABC divides any of its three bisectors into two segments (BI and IP, as we see in the picture above) which are proportional to the sum of the sides (AB and BC) adjacent to the relative angle of the bisector and to the third side (AC):

Incenter theorem formula

Angle Bisector Theorem

Drawing the Bisector Theorem

The angle bisector theorem states than in a triangle Δ ABC the ratio between the length of two sides adjacent to the vertex (side AB and side BC) relative to one of its bisectors (Bb) is equal to the ratio between the corresponding segments where the bisector divides the opposite side (segment AP and segment PC).

In other words, an angle bisector of a triangle divides the opposite side in two segments that are proportional to the other two sides of the triangle.

So, by the angle bisector theorem:

Triangle Bisector Theorem Formula

In addition, but not included in this theorem, it’s also true that:

Calculations in the Triangle Bisector Theorem

Coordinates of the Incenter of a Triangle

We can to locate the coordinates of the incenter I of a triangle Δ ABC if we know the coordinates of its vertices (A, B, and C), and its sides’ lengths (a, b, and c).

Drawing the coordinates of the incenter of a triangle

Note that the coordinates of the incenter I are the weighted average of the coordinates of the vertices, where the weights are the lengths of the corresponding sides. Then, the coordinates of the incenter I is given by the formula:

Formula of the coordinates of the incenter of a triangle

Euler Line

In any non-equilateral triangle the orthocenter (H), the centroid (G) and the circumcenter (O) are aligned. The line that contains these three points is called the Euler line.

Drawing of Euler's line

In an equilateral triangle all three centers are in the same place.

The relative distances between the triangle centers remain constant.

Distances between centers:

It is true that the distance from the orthocenter (H) to the centroid (G) is twice that of the centroid (G) to the circumcenter (O). Or put another way, the HG segment is twice the GO segment:

Formula for the relation of the distances between centers on the Euler line

When the triangle is equilateral, the barycenter, orthocenter, circumcenter, and incenter coincide in the same interior point, which is at the same distance from the three vertices.

This distance to the three vertices of an equilateral triangle is equal to Distance 1 on Euler's Line from one side and, therefore, Distance 1 on Euler's Line to the vertex, being h its altitude (or height).

Exercise 1

Drawing of exercise 1 at the incenter of a triangle

Find the coordinates of the incenter I of a triangle ABC with the vertex coordinates A(3,5), B(4,-1) and C(-4,1).

SOLUTION:

We solve this exercise using an analytical approach. For this, it will be enough to find the equations of two of the angle bisectors. For instance, Ba (bisector line of the internal angle of vertex A) and Bb (that bisects vertex B’s angle).

Finally, we find the point of intersection of both angle bisectors, which it’s the incenter (I) that we are searching for.

Calculation drawing in exercise 1

STEP 1: Find the Equation for the lines of the three sides.

We find the equations of the three lines that pass through the three sides of the triangle Δ ABC. Each one is obtained because we know the coordinates of two points on each line, which are the three vertices. The general equation of the line that passes through two known points is:

Formula of the line that passes through two points in example 1

Firstly, we find the equation of the line that pass through side AB:

Calculation of line AB in example 1

Then, we find the equation of the line passing through side BC.

Calculation of line BC in example 1

Finally, we calculate the equation of the line that pass through side CA.

Calculation of line CA in example 1

STEP 2:

We have the equations of two lines (angle bisectors) that intersect at a point (in this case, at the incenter I):

Calculation of the incenter equations in Example 1

So, the equations of the bisectors of the angles between this two lines are given by:

Calculation of the bisector equation in example 1

Remember that for the triangle in the exercise we have found the three equations, corresponding to the three sides of the triangle Δ ABC.

  • Side AB equation:
    Calculation of the equation for side AB in example 1
  • Side BC equation:
    Calculation of the equation for side BC in example 1
  • Side CA equation:
    Calculation of the equation for side CA in example 1

STEP 3:

Now that we have the equations for the three sides and the angle bisector formula we can find the equations of two of the three angle bisectors of the triangle.

We calculate the angle bisector Ba that divides the angle of the vertex A from the equations of sides AB (6x + y – 23 = 0) and CA (-4x + 7y – 23 = 0):

Calculation of the bisector equation in example 1

Substitute the values,

Calculation of the equation of the bisector Ba in Example 1

Then, we find the angle bisector Bb that divides the angle of the vertex B from the equations of sides AB (6x + y – 23 = 0) and BC (x + 4y = 0).

Again, starting from the formula for the bisector angle:

Calculation of the bisector equation in example 1

Substitute the values,

Calculation of the equation of the bisector Bb in Example 1

For this second equation, the minus sign is taken from ± because the line of the angle bisector Bb has a negative slope.

Graphically, a negative slope means that as the line on the line graph moves from left to right, the line falls.

Well, now we have a system of equations of the first degree with two unknowns corresponding to the equations of the lines of the angle bisectors Ba and Bb:

Calculation of solution 1 in example 1

Subtract member from member of the first equation from the second equation:

Calculation of solution 2 in example 1

Substitute the value of y in either of the two equations:

Calculation of solution 3 in example 1

We have solved the exercise, finding out the coordinates of the incenter, which are I(1.47 , 1.75).

Drawing the solution in exercise 1

Exercise 2

Drawing of exercise 2 at the incenter of a triangle

Find the coordinates of the incenter I of a triangle Δ ABC with the vertex coordinates A (3, 5), B (4, -1) y C (-4, 1), like in the exercise above, but now knowing length’s sides: CB = a = 8.25, CA = b = 8.06 and AB = c = 6.08.

With these given data we directly apply the equations of the coordinates of the incenter previously exposed:

Calculation of the coordinate solution in Example 2

Finally, we obtain the same coordinates of the incenter I for the triangle Δ ABC as those obtained with the procedure of exercise 1, I (1,47 , 1,75).

Exercise 3

In a triangle Δ ABC, let a, b, and c denote the length of sides opposite to vertices A, B, and C respectively. If a = 6 cm, b = 7 cm and c = 9 cm, find the radius r of the inscribed circle whose center is the incenter I, the point where the angle bisectors intersect.

SOLUTION:

STEP 1: Find the semiperimeter.

If the sides have length a, b, c, we define the semiperimeter s to be half their sum, so s = (a+b+c) /2. So, we get that the semiperimeter is:

Calculation of the semiperimeter in Example 2

STEP 2: Find the radius r.

Apply the formula for the inradius r of the inscribed circle (or incircle):

Formula for the inradius inscribed in the triangle with center at the incenter

The inradius is:

Calculation of the inradius in example 3

Therefore the answer is r = 1.91 cm.

Exercise 4

Let a = 4 cm, b = 3 cm and c = 2 cm, be the sides of a triangle Δ ABC.

  1. Drawing 1 in exercise 4

    Find the angle bisectors Ba, Bb and Bc.

  2. Drawing 2 in exercise 4

    Find the radius r of the inscribed circle for the triangle.

SOLUTION:

  1. Firstly we find the semiperimeter: s = 4.5 cm. Now, we can calculate the three angle bisectors calling to the angle bisector formula:
    Formula of angle bisector in example 4

    Remember that if the side lengths of a triangle are a, b and c, the semiperimeter s = (a+b+c) /2, and A is the angle opposite side a, then the length of the internal bisector of angle A. Substitute the values:

    Calculation of the semiperimeter in example 4
  2. We know the semiperimeter s and the sides of the triangle, thus applying the inradius formula we have that:
    Result of the semiperimeter in example 4

AUTHOR: Bernat Requena Serra

YEAR: 2020


IF YOU LIKED IT, SHARE IT!

Leave a Reply

Your email address will not be published.