A triangle is a closed figure enclosed by three line segments. It has three sides, three angles and three vertices. Bisector, as the name implies, divides anything in equal halves. Here, in this case we are referring to line or line segment that divides another line segment in two equal halves. Generally, we have two types of bisectors:

1) Angle Bisectors

2)
Perpendicular Bisectors.

## Bisector of a Triangle Definition

The angle bisector of an angle of a triangle is the line segment that bisects the given angle of the triangle. This angle bisector is always equidistant from the sides of the triangle containing the angle that is bisected. All three angle bisectors of a triangle meet at a common point which is known as the in-center of the triangle.  With the help of this in-center and radius equal to the perpendicular distance between the point and a side, we can draw a circle which is inscribed in the triangle and the sides of the triangle will be tangent to that circle. This circle is known as in-circle.

## Bisector of a Triangle Theorem

The angle bisector of an angle of a triangle divides the opposite side in two parts. The angle bisector theorem states that the two segments are proportional to the other two sides of that triangle.

This implies that if ABC is a triangle and AD is the angle bisector, then according to the angle bisector theorem,

$\frac{AB}{AC}$ = $\frac{BD}{DC}$

This theorem can also be proved easily.

We produce BE || DA such that BE meets CA produced in E.

Now since BE || DA, therefore, $\frac{CD}{DB}$ = $\frac{CA}{AE}$ (Basic proportionality theorem)

Also $\angle4$ = $\angle1$ (corresponding angles)

Also, $\angle1$ = $\angle2$

$\Rightarrow \angle2$ = $\angle4$

Again, $\angle2$ = $\angle3$ (alternate interior angles)

$\Rightarrow \angle4$ = $\angle3$

$\Rightarrow$ Triangle ABE is isosceles

$\Rightarrow$ AB = AE

$\Rightarrow$ $\frac{CD}{DB}$ = $\frac{AC}{AB}$

Hence, theorem proved.

## How to Draw Right Bisector

A right bisector or a perpendicular bisector has two properties:

1) It is perpendicular to the line on which it is drawn.

2) It divides the line segment in two equal halves or simply bisects it.

A perpendicular bisector is very easy to be made. Open the compass more than half the length of the line segment whose perpendicular bisector is to be drawn. Then from one end, draw two arcs on opposite sides of the line segment. Now from another end, we again draw two arcs intersecting the previous arcs. Join the two points obtain. And then verify the construction for bisecting the line segment and also for making right angle with it.

## Right Bisector of a Triangle

Just like angle bisectors of the triangle, all three right bisectors of the triangle also meet at a common point known as circum-center. This circum-center can be used to draw a circle with radius equal to the distance between the circum-center and the vertex of the triangle. This circle will inscribe the triangle within it with the sides of the triangle acting as the chords of the circle. All the vertices of the triangle will lie on the circle. This circle is commonly known as circum-circle.

It is important to note that when the given triangle is an equilateral triangle then the in-center and the circum-center always coincide. Let us see the differences between the in-center and the circum-center of the triangle.

The first one lies in the definition of the two that in-circle is the intersection point of the angle bisectors and the circum-center is the intersection point of the perpendicular bisectors.

The second one is that in-center is equidistant from all the sides, while, circum- center is equidistant from all the vertices.

The third difference is that the in-circle drawn from in- center is inscribed within the triangle while the triangle is inscribed in the circum-circle that is drawn using the circum-center.

## Examples on Bisector of a Triangle

Let us see some examples on the basis of the angle bisector theorem of a triangle.

Example 1: Find x

Solution:

According to the angle bisector theorem we have

$\frac{(4 x + 1)}{15}$ = $\frac{5}{3}$

$\Rightarrow 3 (4 x + 1) = 5 . 15$

$\Rightarrow 12 x + 3 = 75$

$\Rightarrow 12 x = 75 – 3$

$\Rightarrow 12 x = 72$

$\Rightarrow x$ = $\frac{72}{12}$

$\Rightarrow x = 6$

Example 2:  Find AD and AB.

Solution :

According to the angle bisector theorem we have

$\frac{(5 x + 12)}{(8 x)}$ = $\frac{DC}{CB}$

Since DC = CB

Therefore, $\frac{(5 x + 12)}{(8 x)}$  = $1$

$\Rightarrow 5 x + 12 = 8 x$

$\Rightarrow 12 = 8 x – 5 x$

$\Rightarrow 3 x = 12$

$\Rightarrow x = 4$

$\Rightarrow AD = 5 x + 12 = 5 (4) + 12 = 20 + 12 = 32$

And $AB = 8 x = 8 (4) = 32$