Incentre of Triangle with Examples and FAQs

Incentre of Triangle

The incentre is the concurrency point where all the three angle bisectors of a triangle intersect and it lies inside the triangle for all triangles. An angle bisector is a line that divides the angle at the respective vertex equally into two halves.

The incentre is at an equal distance, i.e., it is equidistant from all three sides of the triangle. Hence a circle can be constructed taking this distance as the radius which is the incircle of the triangle. This incircle is the largest circle that can be enclosed in the triangle.

In the figure above G is the incentre and the circle inside is the incircle.

Construction of an Incentre

The incentre of a triangle can be constructed by following the steps given below:

• Draw angle bisectors of all the vertices of the triangle.
• The point of intersection of these three bisectors is the incentre of the triangle.

Construction of Incircle from Incentre

After we determine the incentre of the triangle, we can easily draw a circle by taking the incentre as the centre and the distance of any side(of the triangle) from this point as radius. This gives us the incircle of that particular triangle.

Properties of Incentre

The properties of incentre are:

• The point of intersection of all the angle bisectors in a triangle is its incentre.
• Incentre is the centre of the incircle of the triangle and hence it lies always inside the triangle.

• The sides of the triangles are tangents to the incircle, this proves that the incentre is equidistant from all sides of the triangle. This distance is the radius of the incircle.

In the figure AG = BG = CG

• The angles are bisected hence from the above figure we have:

∠YXB = ∠ZXB; ∠ZYC = ∠XYC; ∠XZA = ∠YZA.

Formula for Incentre

The formula to find the incentre of a triangle if the coordinates of the vertices of the three points are given is:

Let the Triangle be ∆XYZ and the coordinates are \mathrm{X}\left(x_{1}, y_{1}\right), \mathrm{Y}\left(x_{2}, y_{2}\right), \mathrm{Z}\left(x_{3}, y_{3}\right)

The coordinate of the incentre G(x,y) is such that

x=\frac{a x_{1}+b x_{2}+c x_{3}}{a+b+c} \quad \text { and } y=\frac{a y_{1}+b y_{2}+c y_{3}}{a+b+c}

Here, a, b and c are the side lengths that can be determined using the distance formula.

\therefore a =\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}},

b =\sqrt{\left(x_{3}-x_{2}\right)^{2}+\left(y_{3}-y_{2}\right)^{2}}

\text { and } c=\sqrt{\left(x_{3}-x_{1}\right)^{2}+\left(y_{3}-y_{1}\right)^{2}}.

Examples

Example 1: What is the measure of the ∠BCF if O is the incentre and AD, BE and CF are the angle bisectors of ∠BAC, ∠ABC and ∠ACB respectively.

Solution:

From the given triangle we know that O is the incentre and AD, BE and CF are the angle bisectors of ∠BAC, ∠ABC and ∠ACB respectively.

⇒ ∠ABE = ∠CBE, ∠BAD = ∠CAD and ∠ACF = ∠BCF.

Therefore, we get:

∠ABE + ∠CBE = ∠ABC

∠BAD + ∠CAD = ∠BAC

∠ACF + ∠BCF = ∠ACB

From the diagram, we can see that ∠ABE = 25° and ∠BAD = 45°.

Now by angle sum property of a triangle:

∠BAC + ∠ABC + ∠ACB = 180°

⇒ (∠BAD + ∠CAD) + (∠ABE + ∠CBE) + (∠ACF + ∠BCF) = 180°

⇒ 2∠BAD + 2∠ABE + 2∠BCF = 180°

⇒ (2 × 45°)+ (2 × 25°)+2∠BCF =180°

⇒ 90° + 50° + 2∠BCF = 180°

⇒ 2∠BCF = (180 – 90 – 50)° = 40°

\therefore \angle BCF=\frac{40^{\circ}}{2}=20^{\circ}

Example 2: If the coordinates of the vertices of a triangle are:

P(-4, 0), Q(0, 3) and  R(4, 0) then find the incentre’s coordinates. 

Solution:

Let us first determine the length of the sides of the given triangle. Using the distance formula, we have:

The coordinates of incentre are given by the formula:

The coordinate of the incentre G(x,y) is such that

x=\frac{a x_{1}+b x_{2}+c x_{3}}{a+b+c} \text { and } y=\frac{a y_{1}+b y_{2}+c y_{3}}{a+b+c}

Where, a, b and c are the side lengths, from the above calculations we can substitute values of a = 5,  b = 5 and c = 8

x=\frac{a x_{1}+b x_{2}+c x_{3}}{a+b+c} \quad \text { and } y=\frac{a y+b y_{2}+c y_{3}}{a+b+c}

\Rightarrow x=\frac{(5 \times(-4))+(5 \times 0)+(8 \times 4)}{5+5+8} \text { and } y=\frac{(5 \times 0)+(5 \times 3)+(8 \times 0)}{5+5+8}

\Rightarrow x=\frac{(-20)+(0)+(32)}{18} \text { and } y=\frac{(0)+(15)+(0)}{18}

\Rightarrow x=\frac{12}{18}=\frac{2}{3} \quad \text { and } y=\frac{15}{18}=\frac{5}{6}

Hence, the coordinates of the incentre is \left(\frac{2}{3}, \frac{5}{6}\right)