# Incenter of a triangle     (No Ratings Yet) Loading... 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. 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: 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.

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: If is a right triangle this relation between inradius and area is: ## 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): ## Angle 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: In addition, but not included in this theorem, it’s also true that: ## 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). 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: ## 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. 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: 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 from one side and, therefore, to the vertex, being h its altitude (or height).

## Exercise 1 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. 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: Firstly, we find the equation of the line that pass through side AB: Then, we find the equation of the line passing through side BC. Finally, we calculate the equation of the line that pass through side CA. STEP 2:

We have the equations of two lines (angle bisectors) that intersect at a point (in this case, at the incenter I): So, the equations of the bisectors of the angles between this two lines are given by: 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: • Side BC equation: • Side CA equation: 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): Substitute the values, 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: Substitute the values, 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: Subtract member from member of the first equation from the second equation: Substitute the value of y in either of the two equations: We have solved the exercise, finding out the coordinates of the incenter, which are I(1.47 , 1.75). ## Exercise 2 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: 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: STEP 2: Find the radius r.

Apply the formula for the inradius r of the inscribed circle (or incircle):  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. Find the angle bisectors Ba, Bb and Bc.

2. 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: 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: 2. We know the semiperimeter s and the sides of the triangle, thus applying the inradius formula we have that: AUTHOR: Bernat Requena Serra

YEAR: 2020