A ray that divides a given angle into two equal angles is known as an angle bisector. A bisector is a line dividing something into two equal parts. In geometry, a bisector is applied to the line segments and angles.

A line that passes through the midpoint of a line segment is the bisector of the line segment. In this article, let us discuss the definition of an angle bisector, the construction of an angle bisector in a triangle, and solve some examples based on the angle bisector.

Angle Bisector of Triangle Definition

An angle bisector is a ray that divides a given angle into two angles with equal measures. We usually divide an angle in a triangle by a line or ray, which is considered an angle 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 \(\angle A B C=80^{\circ}\). An angle bisector divides it into two equal angles of \(40^{\circ}\).

Construction of Bisector of a Given Angle

To draw a ray \(AX\) bisecting a given angle \(\angle BAC\), 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 \(\angle B A C \).

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

Example

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 vertices \(A, B\) and \(C\) respectively. The angle bisector \(l_{a}\) divides angle \(\alpha\) into two equal parts \(\frac{\alpha}{2}\) 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 \(\frac{\gamma}{2}\).

Properties of Angle Bisector of Triangle

1. Any point on the bisector of an angle is equidistant from the sides of the angle.
2. In a triangle, the angle bisector divides the opposite side in the ratio of the adjacent sides that contain the angle.

Internal and External Bisector of an Angle of Triangle

The bisector of a triangle that divides the opposite side internally in the ratio of corresponding sides containing angles is known as the internal bisector of an angle of a triangle.

The bisector of a triangle that divides the opposite side externally in the ratio of corresponding sides containing angles is known as the external bisector of an angle of a triangle.

Theorems on Internal and External Angle Bisectors of a Triangle

In the below two theorems, we will learn that the internal bisector angle of a triangle divides the opposite side in the ratio of the sides containing the angle and vice-versa.

Theorem 1: 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 \(\Delta ACE,\) we have
\(\angle 3 = \angle 4\)
\( \Rightarrow AE = AC……\left( {iii} \right)\) [Sides opposite to equal angles are equal}
Now, in \(\Delta BCE,\) we have
\(DA||CE\)
\( \Rightarrow \frac{{BD}}{{DC}} = \frac{{BA}}{{AE}}\) [Basic proportionality theorem]

\( \Rightarrow \frac{{BD}}{{DC}} = \frac{{AB}}{{AC}}\) [From \(\left( {iii} \right)\)]

Hence, proved.

Theorem 2: In a triangle \(A B C\), if \(D\) is a point on \(B C\) such that \(\frac{B D}{D C}=\frac{A B}{A C}\), then \(A D\) is the bisector of \(\angle A\).

Given: In \(\triangle A B C\), in which \(D\) is a point on \(BC\) such that, \(\frac{B D}{D C}=\frac{A B}{A C}\)

Construction: Produce \(B A\) to \(E\) such that \(A E=A C\). Join \(E C\).

Proof: In \(\triangle A C E\), we have
\(A E=A C \) [By construction]
\(\Rightarrow \angle 3=\angle 4….(1)\)

Now, \(\frac{B D}{D C}=\frac{A B}{A C}\)
\(\Rightarrow \frac{B D}{D C}=\frac{B A}{A E} \quad[A C=A E]\)

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

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

Therefore, by the converse of the Basic proportionality theorem, we have
\(D A || C E\)
\(\Rightarrow \angle 1=\angle 4 \ldots.(ii)\) [ Corresponding angles ]

and, \(\angle 2=\angle 3…(iii)\) [Alternate angles]

But, \(\angle 3=\angle 4\) from \((i)\)
\(\Rightarrow \angle 1=\angle 2\) [from \((ii)\) and \((iii)\)]

Hence, \(A D\) is the bisector of \(\angle A\).

In the following theorem, we shall prove that the bisector of the exterior of an angle of a triangle divides the opposite side externally in the ratio of the sides containing the angle.

Theorem 3: 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 in \(D\)

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

Construction: Draw \(CE || DA\) meeting \(AB\) in \(E\).

Proof: Since \(CE || DA\) and \(AC\) intersects them.

Therefore, \(\angle 1=\angle 3…(i)\)
Also, \(CE || DA\) and \(BK\) intersects them.
Therefore, \(\angle 2=\angle 4\)
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, \(CE || DA\) and \(BK\) 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.

Solved Examples

Let us understand the concept through some Angle Bisector of Triangle practices problem.

Q.1. In the below figure, \(AD\) is the bisector of \(\angle A\). If \(B D=4 \mathrm{~cm}\) and \(D C=3 \mathrm{~cm}\) Find \(AC\).

Ans: In \(\triangle A B C, A D\) is the bisector of \(\angle A\).
Therefore, \(\frac{B D}{C D}=\frac{A B}{A C}\)
\(\Rightarrow \frac{4}{3}=\frac{6}{A C}\)
\(\Rightarrow 4 A C=18\)
\(\Rightarrow A C=\frac{18}{4}=\frac{9}{2}=4.5 \mathrm{~cm}\)

Q.2. In the below figure, \(AD\) is the bisector of \(\angle B A C\). If \(A B=10 \mathrm{~cm}, A C=14 \mathrm{~cm}\) and \(B C=6 \mathrm{~cm}\), find \(B D\) and \(D C\).
Ans: Let \(B D=x \mathrm{~cm}\). Then, \(D C=(6-x) \mathrm{cm}\)

Since \(AD\) is the bisector of \(\angle A\).
\(\frac{A B}{A C}=\frac{B D}{D C}\)
\(\Rightarrow \frac{10}{14}=\frac{B D}{D C}\)
\(\Rightarrow \frac{10}{14}=\frac{x}{6-x}\)
\(\Rightarrow 30-5 x=7 x\)
\(\Rightarrow 12 x=30\)
\(\Rightarrow x=\frac{5}{2}=2.5 \mathrm{~cm}\)
\(\Rightarrow B D=2.5 \mathrm{~cm}\) and \(D C=6-x=6-2.5=3.5 \mathrm{~cm}\)

Q.3. If the diagonal \(BD\) of a quadrilateral \(ABCD\) bisect both \(\angle B\) and \(\angle D\), show that \(\frac{A B}{A C}=\frac{A D}{C D}\).
Ans: Given: A quadrilateral \(ABCD\) in which the diagonal \(BD\) bisects \(\angle B\) and \(\angle D\).
To prove: \(\frac{A B}{A C}=\frac{A D}{C D}\)
Construction: Join \(AC\) intersecting \(BD\) in \(O\).

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

Q.4. If the bisector of an angle of a triangle bisects the opposite side, prove that the triangle is isosceles.
Ans:

Given: In \(\Delta ABC,\) the bisector \(AD\) of \(\angle A\) bisects the side \(BC.\)
To prove: \(AB = AC\)
Proof: In \(\Delta ABC,\,AD\) is the bisector of \(\angle A.\)
Therefore, \(\frac{{AB}}{{AC}} = \frac{{BD}}{{DC}}\)
\( \Rightarrow \frac{{AB}}{{AC}} = 1\) (\(D\) is the midpoint of \(BC.\) So, \(BD = DC\))
\( \Rightarrow AB = AC\)
Hence, \(\Delta ABC\) is an isosceles triangle.

Q.5. In \(\triangle A B C\), the bisector of \(\angle B\) meets \(AC\) at \(D\). A line \(PQ || AC\) meets \(AB, BC\) and \(B D\) at \(P, Q\) and \(R\) respectively. Show that
(i) \(P R . B Q=Q R . B P\)
(ii) \(A B \times C Q=B C \times A P\)
Ans:
Given: In \(\triangle A B C\), the bisector of \(\angle B\) and line \(PQ ||AC\) meets \(A B, B C\) and \(B D\) at \(P, Q\) and \(R\) respectively.

To prove:
(i) \(P R . B Q=Q R . B P\)
(ii) \(A B \times C Q=B C \times A P\)

Proof:
(i) In \(\triangle B Q P, B R\) is the bisector of \(\angle B\).
Therefore, \(\frac{B Q}{B P}=\frac{Q R}{P R}\)
\(\Rightarrow B Q \cdot P R=B P \cdot Q R\)
\(\Rightarrow P R . B Q=Q R . B P\)
(ii) In \(\triangle A B C\), we have \(PQ || AC\)
\(\Rightarrow \frac{A B}{A P}=\frac{C B}{C Q}\) (Thales theorem)
\(\Rightarrow A B \times C Q=B C \times A P\)
The Angle Bisector of Triangle facts can be understood from the above-solved questions.

Summary

We studied the definition of angle bisectors, construction of angle bisectors, the definition of angle bisector of a triangle, properties of angle bisector of a triangle and theorems on internal and external angle bisectors of a triangle. Also, we solved some example problems based on the angle bisector of a triangle to revise the learned concept.

FAQs

Q.1. Does angle bisector cut an angle in half?
Ans: Yes, the angle bisector cut an angle in half.
Example: Consider an angle \(\angle A B C=60^{\circ}\). An angle bisector divides it into two equal angles of \(30^{\circ}\).

Q.2. Explain the angle bisector of a triangle with an example?
Ans: 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 three angle bisectors of a triangle meet in a single point called the incentre. This point is always inside the triangle.
Example: Consider an angle \(\angle A B C=90^{\circ}\). An angle bisector divides it into two equal angles of \(45^{\circ}\).

Q.3. What is an angle bisector, and what does it do?
Ans: A ray that divides a given angle into two angles with equal measures is called an angle bisector. So, it divides an angle into equal halves.
Example: Consider an angle \(\angle A B C=120^{\circ}\). An angle bisector divides it into two equal angles of \(60^{\circ}\).

Q.4. What is the property of an angle bisector of a triangle?
Ans: The properties of the angle bisector of a triangle are:
1. Any point on the bisector of an angle is equidistant from the sides of the angle.
2. In a triangle, the angle bisector divides the opposite side in the ratio of the adjacent sides.

Q.5. Does the angle bisector go through the midpoint of the side opposite to it?
Ans: A bisector of a line segment will pass through the midpoint of the line segment. To bisect an angle means to divide it into two equal parts. However, an angle bisector need not bisect the side opposite to it. So, we cannot say that the angle bisector will go through the midpoint of the side opposite to it.

Q.6. 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.

We hope the article on Angle Bisector of Triangle will help solve the doubts. Readers can take Angle Bisector of Triangle notes from the article to revise the concept quickly.