Geometry is a branch of mathematics that deals with construction of shapes and problems based on them. In elementary geometry, we learn about bisectors. The term bisector itself indicates something that divides another thing into two equal sections. In geometry, the bisectors that we study are line bisectors, angle bisectors, and perpendicular bisectors. Let us go ahead and learn them in detail.

## Definition

Bisector is defined as a line that divides or separates something into two equal halves. It could divide an angle into two equal parts, and then it would be known as angle bisector. If it divides a line into two equal halves then it shall be called line bisector and if the bisector is perpendicular to the line and makes two right angles at the point of bisecting then it will be called perpendicular bisector.

## Theorem

Angle Bisector Theorem:  In a triangle, the angular bisector cuts the opposite side of the triangle into two parts which are proportional to the rest of the two sides of the triangle.
Line Segment Theorem: If a line segment is cut by another line segment such that it is divided into two equal parts then it is known as the line segment theorem.
Perpendicular Bisector Theorem: If the line bisector cuts the line segment into two equal angles each of 90 degrees, then it is said to be the perpendicular bisector of the line segment.

## Angle Bisector

An angle bisector is defined as the line which cuts the angle into equal angles. If any segment, ray or light passing through the vertex of the angle divides the angles into two congruent angles then it is called the angle bisector.

Angle Bisector Theorems:

The angle bisector theorem states that, in given triangle say $PQR$ and angle bisector $PS$, where $S$ lies on the side $PQ$, then $\frac{q}{m}$ = $\frac{r}{n}$

Prove:

We need to two similar triangles to prove the angle bisector theorem. We see that the given triangle $PQR$ is separated into two smaller triangles $PRS$ and $PQS$ which are not similar to each other. So, we need to draw another triangle which shall be a similar triangle.

To do that, we extend the angle bisector $PS$ and draw a line parallel to $PR$ originating from the vertex $Q$ and intersecting the extended angle bisector $PS$ at $T$ (say).

Let us take up the triangles $PRS$ and $TQS$,

Angle $RPS$ = Angle $QTS$ (Alternate Interior angles)

Angle $RSP$ = Angle $QST$ (Vertically opposite angles)

Therefore, Triangle $PRS$ similar to Triangle $TQS$ (AA postulate)

So, their corresponding sides are in proportion:

$\frac{PR}{RS}$ = $\frac{TQ}{QS}$

$\frac{q}{m}$ = $\frac{u}{n}$ ...... Equation 1

Now, we see that angle $QPS$ = angle $RPS$ (angle bisector)

Angle $QPS$ = Angle $QTS$ (As angle $RPS$ and angle $QTS$ are alternate interior angles)

Let us consider the bigger triangle $PQT$, where angle $QPS$ and angle $QTS$ are the base angles. As the base angles are equal their corresponding sides are also equal forming an isosceles triangle $PQT$. So, $PQ$ = $TQ$ or $r$ = $u$. Plugging in this value of $r$ = $u$ in equation $1$, we get $\frac{q}{m}$ = $\frac{r}{n}$Hence, the angle bisector theorem is proved.

## Line Segment Bisector

### Line segment bisector is a line, line segment or ray that cuts another line segment into two equal segments.

In the above diagram, we see that the line segment $AE$ cuts the line segment $BD$ into two equal parts $BC$ = $CD$.

If line segment $AE$ crosses the line segment $BD$ at a right angle, then it is called the perpendicular bisector of $BD$.

## Perpendicular Bisector

### If a line segment cuts another line segment into two equal halves and making 90 degrees each equal angle at the point of cutting, then it is known as the line bisector.

a) Construction of Perpendicular Bisector

Steps to construct a perpendicular bisector are as follows:

i) We draw a line segment of any desired length.

ii) Take a compass and place its needle on one end of the line segment.

iii) Take the width of the compass slightly greater than the middle point of the line segment.

iv) Draw two arcs with that width of the compass, one above the line segment and the other below the line segment.

v) Without changing the width, place the needle of the compass on the other end of the line segment.

vi) Draw two arcs one above the line segment and another below the line segment intersecting the previous two arcs drawn.

vii) Then placing a scale, draw a straight line passing through the two intersection points.

viii) This line cuts the line segment at a point which is the midpoint of the line segment.

ix) Thus, the line segment which is cutting the other line segment into two equal parts and making 90 degree or right angle at the crossing point is called the line bisector.
b) Proof of Perpendicular Bisector of a Segment:

Given: $AD$ perpendicular $CB$

$AD$ bisects $CB$

Prove: $AC$ = $AB$

To prove it we need to show triangle $ADC$ congruent triangle $ABD$

c) Perpendicular Bisector of a Triangle Theorem:

In a triangle, the perpendicular bisectors of the sides of the triangle meet at a point which is equidistant from the three vertices of the triangle

In the above figure $m1,\ m2$ and $m3$ are the perpendicular bisectors of the sides of the triangle $AB,\ BC,$ and $CA$. Because $G$ lies on the line $m3$ it is equidistant from the points $C$ and $B$. That is, $CG$ = $BG$. Similarly, as the point $G$ lies on the line $m2$ it is equidistant from the vertices $A$ and $B$. Similarly $G$ also lies on the line $m1$ so it is equidistant from the vertices $A$ and $C$. Therefore, we derive $CG$ = $BG$ = $AG$. So, the lines $m1,\ m2,$ and $m3$ are concurrent at a point $G$

## Examples

Example 1: Given in the above figure is that the line segment $BG$ = $6x + 5$ and $GA$ = $35$. Please find out the value of $x$

Solution:

$G$ is the point of concurrency of the three perpendicular bisectors of the triangle $ABC$. Therefore, $G$ being the circumcenter is equidistant from the three vertices of the triangle. We can say,

$BG$ = $CG$

$6x + 5$ = $35$

$6x$ = $35$ - $5$

$6x$ = $30$

$X$ = $\frac{30}{6}$

$X$ = $5$
Example 2: The coordinates of the points $B$ and $D$ are $(2,8)$ and $(-6,-4)$. Please find out the coordinates of the point $C$. Given, $AC$ is the line bisector of $BC$

Solution:

We apply the midpoint formula to find out the point $C$ as $AC$ is the line bisector

$(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$

So, $Cx$ = $\frac{(2-6)}{2}$ = $-2$ and $Cy$ = $\frac{(8-4)}{2}$ = $2$

Therefore, the coordinates of point $C$ is $(-2,2)$.