Angle Bisector Theorem: Proofs and Solved Examples

One of the most fundamental theorems in mathematics, particularly in geometry, is the Angle Bisector Theorem. According to the Angle Bisector Theorem, a triangle’s opposite side will be divided into two proportional segments to the triangle’s other two sides.

Angle bisector theorem applies to all types of triangles, such as equilateral triangles, isosceles triangles, and right-angled triangles, etc. In this article, you will learn in detail about Angle Bisector Theorem, Proofs, Solved Examples. etc. Continue reading to know more.

Introduction to Angle Bisector Theorem

The Angle Bisector Theorem helps to find unknown lengths of sides of triangles because an angle bisector divides the side opposite to that angle into two segments proportional to the triangle’s other two sides.

You are shining a flashlight and would like to know where the strongest light from the flashlight is. Suppose we view the light as an angle, and the angle bisector of that angle intersects the imaginary other side of our triangle. In that case, the Triangle Angle Bisector Theorem says that the angle bisector divides the ‘opposite side’ of the triangle in a proportion equal to the proportion of the other two sides.

That means that the strongest light coming from the flashlight will be where the angle bisector intersects the object you are shining the light at. So, the light will be strongest at the angle bisector and slowly lose power as it goes further towards the sides of the triangle.

Definition and Statement of Angle Bisector Theorem

The Angle Bisector Theorem helps in finding unknown lengths of sides of triangles because an angle bisector divides the side opposite that angle into two segments that are proportional to the triangle’s other two sides.

In other words, it states that if a point is on the angle bisector of an angle in a triangle, then the point is equidistant from the sides of the angle.

Angle Bisector

An angle bisector is a ray or line which divides the given angle into two congruent angles.
The properties of an angle bisector are given below:

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.

Construction of Angle Bisector

Follow the steps given below for constructing an angle bisector with an example

1. Step \(1:\) Draw any angle, say \(\angle ABC\).

2. Step \(2:\) Taking \(B\) as the centre and any appropriate radius, draw an arc to intersect the rays \(BA\) and \(BC\) at, say, \(E\) and \(D,\) respectively.

3. Step \(3:\) Now, taking \(D\) and \(E\) as centres and with a radius more than half of \(DE,\) draw an arc to intersect each other at \(F.\)

4. Step \(4:\) Draw ray \(BF.\) This ray \(BF\) is the required angle bisector of angle \(ABC.\)

Internal Angle Bisector Theorem

An angle bisector of a triangle divides the opposite side into two segments proportional to the other two sides of the triangle. In the triangle \(ABC,\) the angle bisector intersects side \(BC\) at point \(D.\)

As per the angle bisector theorem, the ratio of the line segment \(BD\) to \(DC\) is equal to the ratio of the length of the side \(AB\) to \(AC.\)

\(\frac{{BD}}{{DC}} = \frac{{AB}}{{AC}}\)

Draw a line \(CE\) from point \(C\) parallel to \(AD.\)

The line \(CE\) intersects the extended line \(BA\) of a triangle at \(E.\)

From the figure,\(CE\parallel DA\) and \(AC\) is transversal, and we know that alternate interior angles formed by a transversal are equal.

\(∠DAC=∠ACE\) (Alternate angles)\(……..(1)\)

Similarly, \(CE\parallel DA\) and \(AC\) is transversal. We have
\(∠BAD=∠AEC\) (Corresponding angles)\(……….(2)\)

Given, \(AD\) is the angle bisector of \(A,\)
\(∠BAD=∠DAC…………(3)\)

From \((1), (2)\) and \((3),\) we have 
\(∠ACE=∠AEC\)

Hence, \(ΔACE\) is an isosceles triangle.

So, in \(ΔACE,\) we have \(AE=AC\) (Sides opposite to equal angles are equal) According to the basic proportionality theorem, if a line is drawn parallel to one side of the triangle, then it divides the other two sides in the same ratio.

Now, in ΔBCE we have,\(CE\parallel DA\).

\(\frac{{BD}}{{DC}} = \frac{{AB}}{{AE}}\)

As, \(AE=AC,\) then the ratio becomes

\(\frac{{BD}}{{DC}} = \frac{{AB}}{{AC}}\)

External Angle Bisector Theorem

The external angle bisector of a triangle divides the opposite side externally in the ratio of the sides containing the angle. This condition usually occurs in non-equilateral triangles.

In the triangle \(ABC, AD\) is the internal bisector of \(∠BAC,\) which meets \(BC\) at \(D.\)

According to the angle bisector theorem, the ratio of the line segment \(BD\) to \(DC\) equals the ratio of the length of the side \(AB\) to \(AC.\)

\(\frac{{BD}}{{DC}} = \frac{{AB}}{{AC}}\)

Draw a line \(CE\) from point \(C\) parallel to \(AD.\)

Here, our given triangle is \(ΔABC\) and \(AD\) is the external angle bisector of \(∠CAP.\)

\(CE\) is drawn parallel to \(DA,\) such that it intersects the line \(AB\) of a triangle at \(E.\)

From the figure,\(CE\parallel DA\) and \(AC\) is transversal, and we know that alternate interior angles formed by a transversal are equal.

\(∠ECA=∠CAD\) (Alternate angles)\(……….(1)\)

Similarly, \(CE\parallel DA\) and \(BP\) is transversal. We have
\((∠ECA=∠DAP)\) (Corresponding angles)\(………….(2)\)

Given, \(AD\) is the angle bisector of \(∠CAP,\)
\(∠CAD=∠DAP………..(3)\)

From \((1), (2)\) and \((3),\) we have 
\(∠CEA=∠ECA\)

Hence, \(ΔACE\) is an isosceles triangle.

So, in \(ΔACE,\) we have \(AE=AC\) (Sides opposite to equal angles are equal) 

According to the basic proportionality theorem, if a line is drawn parallel to one side of the triangle, then it divides the other two sides in the same ratio.

Now, in \(ΔBDA\) we have, \(CE\parallel DA\).

So, \(\frac{{BE}}{{EA}} = \frac{{BC}}{{CD}} \Rightarrow \frac{{BE}}{{EA}} + 1 = \frac{{BC}}{{CD}} + 1\)

\( \Rightarrow \frac{{BE + EA}}{{EA}} = \frac{{BC + CD}}{{CD}}\)

\( \Rightarrow \frac{{AB}}{{EA}} = \frac{{BD}}{{CD}}\)

\( \Rightarrow \frac{{AB}}{{AC}} = \frac{{BD}}{{CD}}\)   Since, [\(AE=AC]\)

Hence, proved.

Converse of Angle Bisector Theorem

If a point lies on the interior of an angle and is equidistant from the angle’s sides, then a line from the angle’s vertex through the point bisects the angle.

\(D\) is a point in the interior of angle \(∠BAC.\) If the perpendicular distances \(\left| {DC} \right|\) and \(\left| {DB} \right|\) are equal then, the line \(AD\) is the angle bisector of angle \(∠BAC.\)

\(\angle BAD \cong \angle CAD\)

Perpendicular Bisector Theorem

According to the perpendicular bisector theorem, if a point is equidistant from the endpoints of a line segment in a triangle, then it is on the perpendicular bisector of the line segment. 

Compare \(ΔAPM\) and \(ΔBPM.\) We have:

1. \(AM=BM\)
2. \(PM=PM\) (common)
3. \(\angle AMP = \angle BMP = {90^{\rm{o}}}\)

We see that \(ΔAPM≅ΔBPM\) by the \(SAS\) congruence criterion.
\(PA=PB,\) which means that \(P\) is equidistant from \(A\) and \(B.\)

It states that a perpendicular bisector is drawn from the vertex to the opposite side, then it divides the segment into two congruent parts.

In the above figure,

1. \(MT=NT\)
2. \(MS=NS\)
3. \(MR=NR\)
4. \(MQ=NQ\)

Solved Examples – Angle Bisector Theorem

Q.1. In \(∆ABC,\) if \(AD\) bisects \(BAC.\) Find the value of \(x\)?

Ans: Given that, \(AD\) is the bisector of \(∠A.\)
According to the angle bisector theorem: \(\frac{{AB}}{{AC}} = \frac{{BD}}{{DC}}\)
\( \Rightarrow \frac{x}{{x – 2}} = \frac{{x + 2}}{{x – 1}}\)
\( \Rightarrow x(x – 1) = (x – 2)(x + 2)\)
\( \Rightarrow {x^2} – x = {x^2} – 4\)
\( \Rightarrow – x = – 4\)
\( \Rightarrow x = 4\)

Q.2. Mrudula has drawn a right-angled triangle as shown below, and she asks Cherry to determine the value of \(x.\) Find.

Ans: Given: \(ΔABC\) is a right triangle, right-angled at \(∠B,\) we can say that \(BD\) bisects the angle \(ABC.\) As \(\angle DBC = {45^{\rm{o}}}\) According to the angle bisector theorem: ABBC=ADDC
Substitute \(AB=5, BC=12, AD=3.5,\) and \(DC=x.\)
\( \Rightarrow \frac{5}{{12}} = \frac{{3.5}}{x}\)
\( \Rightarrow 5x = 12 \times 3.5\)
\( \Rightarrow x = \frac{{42}}{5}\)
\( \Rightarrow x = 8.4\)

Q.3. In a pyramid, line segment \(AD\) is the perpendicular bisector of triangle \(ABC\) on line segment \(BC.\) If \(AB=20\) feet and \(BD=7\) feet, find the length of the side \(AC.\)

Ans: Given: \(AD\) is the perpendicular bisector on the line segment \(BC.\)
By Perpendicular bisector theorem, any point on line segment \(AD\) is at an equal distance from points \(B\) and \(C.\)
Therefore, \(AB=AC\)
From the figure, \(AB=20\) feet.
So, \(AC=20\) feet.

Q.4. In \(∆ABC,\) if \(AD\) bisects \(∠A.\) Find the value of \(x\)?

Ans: Given that, \(AD\) is the bisector of \(∠A.\)
According to the angle bisector theorem: \(\frac{{AB}}{{AC}} = \frac{{BD}}{{DC}}\)
\( \Rightarrow \frac{6}{x} = \frac{{12}}{{18}}\)
\( \Rightarrow \frac{6}{x} = \frac{2}{3}\)
\( \Rightarrow 2x = 18\)
\( \Rightarrow x = \frac{{18}}{2}\)
\( \Rightarrow x = 9\)

Q.5. Here, \(WX⊥ZY,\) Find \(x\) and the length of each segment.

Ans: Given:
By using the Perpendicular bisector theorem \(WZ=WY.\)
\(⇒2x+11=4x-5\)
\(⇒4x-2x=11+5\)
\(⇒2x=16\)
\( \Rightarrow x = \frac{{16}}{2} = 8\)
So, the values of \(x\) is 8the sides \(WZ=28+11=27\) and \(WY=48-5=27.\)

Summary

The angle bisector of any angle will divide the opposite side in the ratio of the sides containing the angle. In this article, we have discussed the most important theorems on the angle bisector. 

We have learnt about the angle bisector theorem proof, angle bisector theorem examples, triangle angle bisector theorem, perpendicular angle bisector theorem, how to construct angle bisector, and other exciting properties and facts around angle bisectors.

Frequently Asked Questions

Q.1. What is an example of a perpendicular bisector?
Ans: The median of a triangle is the line that joins the vertex of the triangle to the midpoint of the opposite side of the vertex. The median of an equilateral triangle is an example of a perpendicular bisector.

Q.2. What is an angle bisector of a triangle?
Ans: An angle bisector is nothing but a ray that divides an angle into two congruent parts. Thus, the ray is called an angle bisector.

Q.3. What are the uses of the angle bisector theorem?
Ans: The angle bisector theorem is commonly used when the angle bisectors and side lengths are known.
An immediate consequence of the theorem is that the angle bisector of the vertex angle of an isosceles triangle will also bisect the opposite side.

Q.4. Explain the angle bisector theorem?
Ans: The angle bisector theorem states that if a ray bisects an angle of a triangle, it divides the opposite side into two parts with an equal ratio to the other two sides.

Q.5. How to implement the angle bisector theorem?
Ans: In the triangle, \(ABC,\) the angle bisector from \(A\) intersects side \(BC\) at point \(D.\)
According to the angle bisector theorem, the ratio of the line segment \(BD\) to \(DC\) equals the ratio of the length of the side \(AB\) to \(AC.\)
\(\frac{{BD}}{{DC}} = \frac{{AB}}{{AC}}\)