Angular Bisectors: Definition, Theorem, Types, Properties, Examples

Angular Bisectors: In geometry, an angle bisector is a line that divides an angle into two equal angles. The term “bisector” refers to something that divides a form or object into two equal halves. An angle bisector is defined as a ray that divides an angle into two equal pieces of the same measure.

Let us quickly review the many types of angles in mathematics before discussing an angle bisector. An angle can be acute (less than 90 degrees, such as a 60-degree angle), obtuse (greater than 90 degrees), or right depending on the inclination between the two arms (exactly 90-degrees). Angle construction is an important part of geometry since it is used to construct other geometric forms, particularly triangles. By simply bisecting several common angles, a range of angles may be created. On this page let us discuss everything about angle bisector. Read further to find more.

Definition of Angle Bisectors

A line that divides a line or an angle into two equal parts is known as a bisector. The bisector of a line segment always contains the midpoint of the segment.

There are two types of bisectors, they are,

1. Perpendicular Bisectors
2. Angular Bisectors

In this article, we will learn about angular bisectors with some examples.

Definition of Angular Bisector

A ray that divides a given angle into two angles with equal measures is called an angular bisector. The word bisector or bisection means dividing one line or something into two equal parts. For example, a line or ray that divide an angle is usually considered an angle bisector in geometry. 

Bisecting an angle means drawing a ray in the interior of the angle, with its initial point at the vertex of the angle such that it divides the angle into two equal parts.

Example: Consider an angle \(\angle A B C=90^{\circ}\). An angle bisector divides it into two equal angles of \(45^{\circ}\).

Construction of the Bisector of a Given Angle

To draw a ray \(A X\) bisecting a given angle \(\angle B A C\), we follow the below steps.

1. With centre \(A\) and any convenient radius, draw an arc cutting \(AB\) and \(AC\) at \(P\) and \(Q\), respectively.
2. With centre \(P\) and radius more than \(\frac{1}{2} P Q\), draw an arc.
3. With centre \(Q\) and the same radius, as in the above step, draw another arc intersecting the arc in step \(2\) at \(R\).
4. Join \(A R\) and produce it to any point \(X\). The ray \(A X\) is the required bisector of \(\angle B A C\).

Verification: Measure \(\angle B A X\) and \(\angle C A X\). We will find that \(\angle B A X=\angle C A X\).

Uses of Angular Bisectors

We can construct angular bisectors in the below polygons.

1. Triangle
2. Quadrilateral

Angular Bisector of a Triangle

A line segment that bisects one of the vertex angles of a triangle and ends up on the corresponding side of a triangle is known as the angle bisector of a triangle.

There are three angle bisectors in a triangle. The bisectors of a triangle meet in a single point called the incentre. This point is always inside the triangle.

In the above triangle \(A B C, \alpha, \beta, \gamma\) are the angles of \(A, B\) and \(C\) respectively. The angle bisector \(l_{a}\) divides angle \(\alpha\) into two equal parts \(\frac{a}{2}\) and \(\frac{\alpha}{2}\), bisector \(l_{b}\) divides angle \(\beta\) into two equal parts \(\frac{\beta}{2}\) and \(\frac{\beta}{2}\) respectively, and bisector \(l_{c}\) divides the angle \(\gamma\) into two equal parts \(\frac{\gamma}{2}\) and \(\frac{\gamma}{2}\).

Perpendicular Bisector of a Triangle

A perpendicular bisector of a triangle is a line passing through the midpoint of each side perpendicular to the given side.

The three perpendicular bisectors of the sides of a triangle meet at one point, called the circumcenter. A point where three or more lines intersect is called a point of coincidence. So, the circumcenter is the point of coincidence of perpendicular bisectors of a triangle. It is equidistant from the vertices of a triangle. That is \(A O=B O=C O\).

Angular Bisector of Quadrilateral

A line segment that bisects one of the vertex angles of a quadrilateral and ends up on the corresponding side of a quadrilateral is known as the angle bisector of a quadrilateral.

There are four angle bisectors in a quadrilateral. The bisectors of a quadrilateral meeting in a single point called the incentre. This point is always inside the quadrilateral.

In the above quadrilateral \(ABCD\), the purple lines are the angle bisectors of each of the internal angles of the quadrilateral. These angle bisectors intersect at the points \(W, X, Y\) and \(Z\). The circle in blue colour is the circumcircle of the quadrilateral \(WXYZ\).

Angular Bisector Theorems of a Triangle

Internal Angle Bisector Theorem Proof: The internal angle bisector of a triangle divides the opposite side internally in the ratio of the sides containing the angle.

Given: In \(\triangle A B C, A D\) is the internal bisector of \(\angle A\) and meets \(B C\) in \(D\).

To prove: \(\frac{B D}{D C}=\frac{A B}{A C}\)

Construction: Draw \(C E || D A\) to meet \(B A\) produced to \(E\).

Proof: \(C E || D A\) and \(A C\) cuts them.

Therefore, \(\angle 2=\angle 3….(i)\) (Alternate angles)

and \(\angle 1=\angle 4….(ii)\) (Corresponding angles)

But, \(\angle 1=\angle 2\) (\(A D\) is the bisector of \(\angle A\))

From \((i)\) and \((ii)\), we get

\(\angle 3=\angle 4\)

Thus, in \(\triangle A C E\), we have

\(\angle 3=\angle 4\)

\(\Rightarrow A E=A C….(iii)\) [ Sides opposite to equal angles are equal]

Now, in \(\triangle B C E\), we have

\(D A || C E\)

\(\Rightarrow \frac{B D}{D C}=\frac{B A}{A E}\) [Basic proportionality theorem]

\(\Rightarrow \frac{B D}{D C}=\frac{A B}{A C}\) [From \((iii)\)]

Hence, proved.

External Angle Bisector Theorem Proof: The external bisector of an angle of a triangle divides the opposite side externally in the ratio of the sides containing the angle.

Given: In \(\triangle A B C, A D\) is the bisector of exterior angle \(\angle A\) and intersects \(B C\) produced to \(D\)

To prove: \(\frac{B D}{C D}=\frac{A B}{A C}\)

Construction: Draw \(C E || D A\) meeting \(A B\) at \(E\).

Proof: \(C E || D A\) and \(A C\) intersects them.

Therefore, \(\angle 1=\angle 3….(i)\) (Alternate angles)

Also, \(C E || D A\) and \(B K\) intersects them.

Therefore, \(\angle 2=\angle 4……(ii)\) (Corresponding angles)

But, \(\angle 1=\angle 2\) [\(A D\) is the bisector of \(\angle C A K\)]

Thus, in \(\triangle A C E\), we have

\(\angle 3=\angle 4\)

\(\Rightarrow A E=A C\) [Sides opposite to equal angles in a triangle are equal] \(….(iii)\)

Also, \(C E || D A\) and \(B K\) intersects them.

Now, in \(\triangle B A D\), we have

\(E C || A D\)

Therefore, \(\frac{B D}{C D}=\frac{B A}{E A}\) [Using corollary of Basic proportionality theorem]

\(\Rightarrow \frac{B D}{C D}=\frac{A B}{A E}\)

\(\Rightarrow \frac{B D}{C D}=\frac{A B}{A C}\) [\(A E=A C\) from \((iii)\)]

Hence, proved.

Angular Bisector Theorem of a Quadrilateral

The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic.

Given: A cyclic quadrilateral \(A B C D\) in which \(A P, B P, C R\) and \(D R\) are the bisectors of \(\angle A, \angle B, \angle C\) and \(\angle D\) respectively, such that a quadrilateral \(P Q R S\) is formed.

To prove: \(P Q R S\) is a cyclic quadrilateral.

Proof: To prove that \(P Q R S\) is a cyclic quadrilateral, it is sufficient to show that \(\angle A P B+\angle C R D=180^{\circ}\)

Since the sum of the angles of a triangle is \(180^{\circ}\), in triangles \(\triangle A P B\) and \(\triangle C R D\), we have

\(\angle A P B+\angle P A B+\angle P B A=180^{\circ}\)

and, \(\angle C R D+\angle R C D+\angle R D C=180^{\circ}\)

\(\Rightarrow \angle A P B+\frac{1}{2} \angle A+\frac{1}{2} \angle B=180^{\circ}[A P\) and \(B P\) are bisectors of \(\angle A\) and \(\angle B\) respectively]

and, \(\angle C R D+\frac{1}{2} \angle C+\frac{1}{2} \angle D=180^{\circ}[C R\) and \(D R\) are bisectors of \(\angle C\) and \(\angle D\) respectively]

Adding above two equations, we get

\(\angle A P B+\frac{1}{2} \angle A+\frac{1}{2} \angle B+\angle C R D+\frac{1}{2} \angle C+\frac{1}{2} \angle D=180^{\circ}+180^{\circ}\)

\(\Rightarrow \angle A P B+\angle C R D+\frac{1}{2}(\angle A+\angle B+\angle C+\angle D)=360^{\circ}\)

\(\Rightarrow \angle A P B+\angle C R D+\frac{1}{2}[(\angle A+\angle C)+(\angle B+\angle D)]=360^{\circ}\)

\(\Rightarrow \angle A P B+\angle C R D+\frac{1}{2}\left(180^{\circ}+180^{\circ}\right)=360^{\circ}\)[\(A B C D\) is a cyclic quadrilateral ]

\(\Rightarrow \angle A P B+\angle C R D=180^{\circ}\)

Hence, \(P Q R S\) is a cyclic quadrilateral.

Solved Examples – Angular Bisectors

Q.1. In \(\triangle A B C, A D\) is the bisector of \(\angle A\) meeting side \(B C\) at \(D\). If \(A D=5.6 \mathrm{~cm}\), \(B C=6 \mathrm{~cm}\) and \(B D=3.2 \mathrm{~cm}\) find \(A C\).
Ans: In \(\triangle A B C, A D\) is the bisector of \(\angle A\).
We know that the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the containing angle.
Therefore, \(\frac{A B}{A C}=\frac{B D}{D C}\)
\(\Rightarrow \frac{5.6}{A C}=\frac{3.2}{6-3.2} \quad[D C=B C-B D]\)
\(\Rightarrow A C=\frac{5.6 \times 2.8}{3.2}\)
\(\Rightarrow A C=0.7 \times 7\)
\(\Rightarrow A C=4.9 \mathrm{~cm}\)

Q.2. In the given figure, \(A E\) is the bisector of the exterior \(\angle C A D\) meeting \(B C\) produced in \(E\). If \(A B=10 \mathrm{~cm}, A C=6 \mathrm{~cm}\) and \(B C=12 \mathrm{~cm}\), find \(C E\).

Answer: By exterior angle bisector theorem, the external bisector of an angle of a triangle divides the opposite side externally in the ratio of the sides containing the angle.
\(\Rightarrow \frac{B E}{C E}=\frac{A B}{A C}\)
\(\Rightarrow \frac{B C+C E}{C E}=\frac{A B}{A C}\)
\(\Rightarrow \frac{12+x}{x}=\frac{5}{3}\)
\(\Rightarrow 36+3 x=5 x\)
\(\Rightarrow 36=2 x\)
\(\Rightarrow x=18 \mathrm{~cm}\)
Therefore, \(C E=18 \mathrm{~cm}\)

Q.3. In the given figure \(\triangle A B C\) is a triangle such that, \(\frac{A B}{A C}=\frac{B D}{D C}\), \(\angle B=70^{\circ}, \angle C=50^{\circ}\). Find \(\angle B A D\).

Ans: We know that if a line through one of the vertex of a triangle divides the opposite side in the ratio of the other two sides, the line bisects the angle at the vertex.
Given: \(\frac{A B}{A C}=\frac{B D}{D C}, \angle B=70^{\circ}, \angle C=50^{\circ}\)
\(\frac{A B}{A C}=\frac{B D}{D C} \Rightarrow A D\) bisects \(\angle A\)
\(\Rightarrow \angle A=180^{\circ}-\left(70^{\circ}+50^{\circ}\right)\)
\(\Rightarrow \angle A=60^{\circ}\)
Therefore, \(\angle B A D=\frac{\angle A}{2}=\frac{60^{\circ}}{2}=30^{\circ}\)

Q.4. \(A B C D\) is a quadrilateral in which \(A B=A D\). The bisectors of \(\angle B A C\) and \(\angle C A D\) intersect the sides \(B C\) and \(C D\) at the points \(E\) and \(F\) respectively. Prove that \(E F || B D\).
Ans:
Given: \(A B C D\) is a quadrilateral in which \(A B=A D\). The bisectors of \(\angle B A C\) and \(\angle C A D\) intersect the sides \(B C\) and \(C D\) at the points \(E\) and \(F\) respectively
To prove: \(E F || B D\)
Construction: Join \(A C, B D\) and \(E F\)

Proof: In \(\triangle C A B, A E\) is the bisector of \(\angle B A C\).
Therefore, \(\frac{A C}{A B}=\frac{C E}{B E}……(i)\)
In \(\triangle A C D, A F\) is the bisector of \(\angle C A D\).
Therefore, \(\frac{A C}{A D}=\frac{C F}{D F}\)
\(\Rightarrow \frac{A C}{A B}=\frac{C F}{D F} [A D=A B]……(ii)\)
From \((i)\) and \((ii)\), we get
\(\frac{C E}{B E}=\frac{C F}{D F}\)
\(\Rightarrow \frac{C E}{E B}=\frac{C F}{F D}\)
Thus, in \(\triangle C B D, E\) and \(F\) divide’ the sides \(C B\) and \(C D\), respectively, in the same ratio. Therefore, by the converse of Thales theorem, we have \(E F || B D\).

Q.5. If the diagonal \(B D\) of a quadrilateral \(A B C D\) bisects both \(\angle B\) and \(\angle D\), show that \(\frac{A B}{B C}=\frac{A D}{C D}\).
Ans:
Given: A quadrilateral \(A B C D\) in which the diagonal \(B D\) bisects \(\angle B\) and \(\angle D\).
To prove: \(\frac{A B}{B C}=\frac{A D}{C D}\)
Construction: Join \(A C\) intersecting \(B D\) at \(O\).

Proof: In \(\triangle A B C, B O\) is the bisector of \(\angle B\).
Therefore, \(\frac{A O}{O C}=\frac{B A}{B C}\)
\(\Rightarrow \frac{O A}{O C}=\frac{A B}{B C}\ldots..(i)\)
In \(\triangle A D C, D O\) is the bisector of \(\angle D\).
Therefore, \(\frac{A O}{O C}=\frac{D A}{D C}\)
\(\Rightarrow \frac{O A}{O C}=\frac{A D}{C D}\ldots.({ii})\)
From \((i)\) and \((ii)\), we get
\(\frac{A B}{B C}=\frac{A D}{C D}\)

Summary

An angle bisector exists for every angle. It’s also the symmetry line that connects the two arms of an angle, allowing you to build smaller angles using it. Let us just say you have to build a 30° angle. By making a 60° angle and then bisecting it, you may do this. Similarly, an angle bisector is used to build 90-degree, 45-degree, 15-degree, and other angles.

In the above article, we learned the definition of angular bisectors, construction of angular bisectors, properties of angular bisector and theorems on internal and external angle bisectors of a triangle, and angular bisector theorem of a quadrilateral. Also, we solved some problems based on the angular bisector of a triangle and quadrilateral.

Frequently Asked Questions (FAQ) – Angular Bisector

Q.1. How do you find the angle bisector?
Ans: A ray that divides a given angle into two angles with equal measures is called an angular bisector.
Bisecting an angle means drawing a ray in the interior of the angle, with its initial point at the vertex of the angle such that it divides the angle into two equal parts.
Example: Consider an angle \(\angle A B C=60^{\circ}\). An angle bisector divides it into two equal angles of \(30^{\circ}\).

Q.2. How do you draw an angular bisector?
Ans: To draw a ray \(A X\) bisecting a given angle \(\angle B A C\), we follow the below steps.
1. With centre \(A\) and any convenient radius, draw an arc cutting \(AB\) and \(AC\) at \(P\) and \(Q\), respectively.
2. With centre \(P\) and radius more than \(\frac{1}{2} P Q\), draw an arc.
3. With centre \(Q\) and the same radius, as in the above step, draw another arc intersecting the arc in step \(2\) at \(R\).
4. Join \(AR\) and produce it to any point \(X\). The ray \(AX\) is the required bisector of \(\angle B A C\).

Q.3. Can you use the angle bisector theorem to solve the problems based on cyclic quadrilateral?
Ans: Yes, the angular bisector theorem of a quadrilateral can solve the problems based on cyclic quadrilateral.
The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic.

Q.4. State the internal angular bisector theorem.
Ans: The internal angle bisector theorem states that the internal angle bisector of a triangle divides the opposite side internally in the ratio of the sides containing the angle.

Q.5. Which is the best definition for angle bisector?
Ans: The line that passes through the vertex of an angle dividing it into two equal parts is known as the angle bisector.

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