Table of Contents

Last modified on September 6th, 2022

The incenter of a triangle is the point where the three interior angle bisectors intersect. The three angle bisectors are always concurrent and always meet in the triangle’s interior. The incenter is thus one of the triangle’s points of concurrency along with the orthocenter, circumcenter, and centroid. It is typically represented by the letter ‘I’.

When a circle is inscribed in a triangle such that the circle touches each side of the triangle, the center of the circle is also called the incenter.

Shown below is a ΔABC with an inscribed circle.

- Found inside a triangle; in ΔABC, ‘I’ is the centroid
- It is the intersection point of three angle bisectors; in ΔABC, AQ, BR, and CP are the three angle bisectors that meet at the triangle incenter ‘I’
- It is the center of the triangle’s incircle

Using the properties of incenter, let us solve an example.

**Find the value of x. Given that ‘I’ is the incenter of the ΔABC.**

Solution:

Given: I is the incenter of the ΔABC

Thus, AI, BI, and CI are the angle bisectors.

From, angle sum theorem, we know,

∠BAI + ∠CBI + ∠ACI = 180°

Now, we can write,

∠BAI + ∠CBI + ∠ACI = 180°/2 (since, ∠BAI, ∠CBI, and ∠ACI are angle bisectors of ∠A, ∠B, and ∠C)

40° + 20° + x = 90°

x = 90° – (40° + 20°)

x = 30°

The easiest way to find the incenter is by determining the inradius or the radius of the incircle. The inradius is denoted by the letter ‘r’. Once the inradius is known, each side length can be determined by the length of the inradius, and the intersection of the three lines will be the incenter. This is possible using coordinate geometry.

Although, the most widely used procedure is through the use of the Cartesian coordinate system described below:

Let ABC be a triangle whose three vertices are (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}) and AD, BE, and CF are the internal bisectors of ∠A, ∠B, and ∠C.

Again, if a, b, and c is the side lengths opposite vertex A, B and C respectively, such that AB = c, BC = a, and CA = b, then, the formula to find the centroid of the triangle is given below:

Let us solve an example to understand the concept better.

**A triangle has vertices at A = (0, 0), B = (14, 0), and C = (5, 12). What are the coordinates of the incenter?**

Solution:

Let the vertices be:

A = (0, 0) =(x_{1}, y_{1})

B = (14, 0) =(x_{2}, y_{2})

C = (5, 12)=(x_{3}, y_{3})

Here, we will use the distance formula to find the side lengths a, b and c, which is given below:

Distance (d) = √(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}

Now,

BC = a = √(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}, herex_{1} = 0, x_{2} = 14, y_{1} = 0, y_{2} = 0

a = √(14 – 5)^{2} + ( 12 – 0)^{2}

a = √81 + 144

a = 15

AC = b = √(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}, herex_{1} = 0, x_{2} = 5, y_{1} = 0, y_{2} = 12

b = √(5 – 0)^{2} + (12 – 0)^{2}

b = √25 + 144

b = 13

AB = c = √(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}, herex_{1} = 0, x_{2} = 14, y_{1} = 0, y_{2} = 0

c = √(14 – 0)^{2} + (0 – 0)^{2}

c = 14

Now, as we know,

Incenter (I) of a triangle = (ax_{1} + bx_{2} + cx_{3}/a + b + c, ay_{1} + by_{2} + cy_{3}/a + b + c), here vertices A = (0, 0), B = (14, 0), and C = (5, 12)and side lengths BC = 15, AC = 13, & AB = 14

I = (15 × 0 + 13 × 14 + 14 × 5/13 + 14 + 15, 15 × 0 + 13 × 0 + 14 × 12/13 + 14 + 15)

I = (15 + 182 + 70/42, 0 + 0 + 168/42)

I = (6, 4)

Thus, the incenter of the triangle lies at (6, 4)**Relation with Inradius and Area**

For a triangle with semiperimeter (s), where s = a + b + c/2 (a, b and c are the side lengths) and inradius (r), the area of the triangle is determined using the formula given below:

Area (A) = s × r

**Find the area of the triangle with sides 3 cm, 4 cm, and 5 cm and inradius 2 cm**

Solution:

**As we know,**

Semiperimeter (s) = a + b + c/2, here a = 3 cm, b = 4 cm, and c = 5 cm

= (3 + 4 + 5)/2 = 6 cm

Now,

Area (A) = = s × r, here s = 6 cm, r = 2 cm

= 6 × 2 = 12 cm^{2}

To construct the incenter of a given triangle one needs to follow the following steps given below.

**Steps:**

- We start with the given triangle.
- Place the compass pointer on any of the three triangle vertices. Adjust the compass to a medium width setting (the exact width is not important).
- Without changing the compass width strike an arc across each of the two adjacent sides.
- Slightly change the compass width (if desired, not compulsory) and from the point each arc crosses the side, draw two arcs inside the triangle, so that they cross each other (using the same compass width for both).
- Using the straightedge draw a line from the vertex of the triangle to where the last two arcs cross.
- Repeat all the above steps for the other two vertices and draw lines as described above. You will now have three new lines.
- Mark a point where three new lines intersect. This is the incenter of the triangle.

**Remember**: You will get the incenter using only two vertices also. The third vertex is optional.

Last modified on September 6th, 2022