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The **angle bisector theorem** highlights the relationship shared between the line segments and sides of a given triangle. Since this theorem applies to all types of triangles, this opens a wide range of word problems, theorems and other applications in geometry.

*The angle bisector theorem shows how the line segments formed by the angle bisector and the sides of the triangle are proportional to each other. *

Thanks to triangle theorems like this, **we can study how smaller triangles within a larger triangle behave**. Learn the basics of the angle bisector theorem, understand its origin, and feel confident when applying the theorem!

## What Is the Angle Bisector Theorem?

The angle bisector theorem is a theorem stating that **when an angle bisector bisects a triangle’s interior angle and divides the angle’s opposite side into two line segments, the following ratios are equal**: *each of the sides includes the angle being bisected and over the length of the adjacent line segment of the opposite side.*

To better understand the angle bisector theorem, take a look at $\Delta ABC$. The angle bisector, $\overline{CO}$, **divides** $\angle ACB$ **into two congruent angles**.

This also results in dividing the opposite side **into two line segments**: $\overline{AB}$. According to the angle bisector theorem, the ratios of the line segments $\overline{AO}$ and $\overline{OB}$ and the triangle’s sides $\overline{AC}$ and $\overline{BC}$ are proportional.

\begin{aligned}\color{DarkOrange}\textbf{Angle Bisec} &\color{DarkOrange}\textbf{tor Theorem}\\\dfrac{\overline{AC}}{\overline{AO}} &=\dfrac{\overline{BC}}{\overline{BO}}\\\dfrac{m}{a} &=\dfrac{n}{b}\end{aligned}

Let’s extend our understanding of the angle bisector theorem by applying what we’ve learned to analyze the triangle shown below. The line segment $\overline{CO}$ divides the angle $\angle ACB$ into two congruent angles, $\angle ACO =\angle OCB =40^{\circ}$. This means that $\overline{CO}$ **is the angle bisector of the angle** $\angle ACB$. The same line segment divides the opposite side, $\overline{AB}$, into two line segments.

The angle bisector theorem states that when this happens, the affected line segments and the two sides of the triangle **are proportional**.

\begin{aligned}\dfrac{AC}{AO} &= \dfrac{BC}{BO}\\\dfrac{24}{18} &= \dfrac{16}{12}\\\dfrac{4}{3} &\overset{\checkmark}{=} \dfrac{4}{3}\end{aligned}

This example highlights the important components needed to apply the angle bisector theorem. It’s now time to understand **how this theorem was established to know it by heart**.

### Proving the Angle Bisector Theorem

When proving the angle bisector theorem, **use the properties of parallel lines and the side splitter theorem**. Begin the setup by extending the side of the triangle then constructing a line that’s parallel to the given angle bisector. These two new lines should meet and form an adjacent triangle.

Take a look at the triangle $\Delta ABC$. It has an angle bisector, $\overline{CO}$, dividing $\angle ACB$ into two congruent angles. **Extend** $AC$ **to form the line segment** $\overline{AP}$ and **construct a line parallel to** $\overline{CO}$ **that meets at** $P$.

We’ve established that $\overline{CO}$ bisects $\angle ACB$, so we have $\angle ACO = \angle OCB$ or $\angle 1 = \angle 2$. Since $\overline{CO}$ is parallel to $\overline{BP}$,** we can relate** $\angle 1$ **and** $\angle 3$ **as well as** $\angle 2$ **and** $\angle 4$:

- The angles $\angle 1$ and $\angle 3$ are corresponding angles, so $\angle 1 = \angle 3$.
- Similarly, since the angles $\angle 2$ and $\angle 4$ are alternate interior angles, $\angle 2 = \angle 4$.

\begin{aligned}\angle 1&= \angle 2\\ \angle 2 &= \angle 4\\\angle 1&= \angle 3\\\\\therefore \angle 3 &= 4\end{aligned}

Looking at the larger triangle $\Delta ABP$, $\overline{CO}$ passes through two sides of the triangle and **the angle bisector is parallel to the third side**, $\overline{BP}$.

Using the side splitter theorem, *the line segments share the following proportionality:*

\begin{aligned}\dfrac{AO}{OB} &= \dfrac{AC}{CP}\end{aligned}

Since $\angle 3 = \angle 4$, **the triangle** $\Delta CBP$ **is isosceles and consequently**, $\overline{CP} = \overline{CB}$. Substitute $\overline {CP}$ with $\overline{CB}$ and *have the following relationship instead:*

\begin{aligned}\dfrac{AO}{OB} &= \dfrac{AC}{CB}\\ \dfrac{AC}{AO} &= \dfrac{CB}{OB}\end{aligned}

This proves that when the angle bisector divides the third side into two line segments, **the sides and the resulting line segments are proportional to each other**.

Now that we’ve proven the angle bisector theorem, it’s time to learn how to apply this theorem to solve different problems involving angle bisectors.

## How To Find the Angle Bisector?

To find the angle bisector of a triangle, apply the converse of the angle bisector theorem by **observing the proportions of the pairs of sides to confirm that the given line segment is an angle bisector**.

*The converse statement establishes that when:*

- The line segment divides a vertex and angle of a triangle.
- It also divides the triangle into smaller triangles with proportional sides.
- The line segment is the angle bisector of the triangle.

This means that when $\overline{CO}$ divides the triangle $\Delta ABC$ into two triangles where the two sides are proportional as shown below, **the line** $\overline{CO}$ **is an angle bisector of** $\angle ACB$.

\begin{aligned}\overline{CO} \text{ divides } &\text{the triangle},\\\dfrac{m}{a}&= \dfrac{n}{b},\\\therefore \overline{CO} \text{ is an an}&\text{gle bisector}\end{aligned}

To confirm that the line $\overline{CO}$ is the angle bisector of $\angle ACB$, **take a look at the ratios of the following line segments and sides of the triangle**: $\overline{AC}$ and $\overline{AO}$ as well as $\overline{CB}$ and $\overline{OB}$.

\begin{aligned}\dfrac{AC}{AO} &= \dfrac{12}{10}\\&= \dfrac{6}{5}\end{aligned} | \begin{aligned}\dfrac{CB}{OB}&= \dfrac{18}{15}\\&=\dfrac{6}{5}\end{aligned} |

\begin{aligned}\dfrac{AC}{AO} &= \dfrac{CB}{OB}\\\Rightarrow \overline{CO}&: \text{Angle Bisector}\end{aligned} |

Using the converse of the angle bisector theorem, **the line segment** $\overline{CO}$ **is indeed the angle bisector of** $\angle ACB$.

*Excited to try out more problems? *

Don’t worry, the section below offers more exercises and practice problems!

### Example 1

In the triangle $\Delta LMN$ the line $\overline{MO}$ bisects $\angle LMO$. Suppose that $\overline{LM} = 20$ cm, $\overline{MN} = 24$ cm, and $\overline{LO} = 15$ cm, what is the length of line segment $\overline{ON}$?

__Solution__

First, **construct a triangle with an angle bisector dividing the angle’s opposite side**. Assign the given lengths of the triangle’s sides and the line segment $\overline{LO}$ as shown below. Let $x$ represent the measure of $\overline{ON}$.

Since $\overline{MO}$ bisects $\angle LMN$ into two congruent angles and using the angle bisector theorem,* the ratios of the sides are as follows:*

\begin{aligned}\dfrac{LM}{LO} &= \dfrac{MN}{ON}\\\dfrac{20}{15} &= \dfrac{24}{x}\end{aligned}

Simplify the equation then** solve** $x$** to find the measure of the line segment** $\overline{ON}$.

\begin{aligned}\dfrac{4}{3} &= \dfrac{24}{x}\\4x&= 24(3)\\4x&= 72\\ x&= 18\end{aligned}

This means that $\overline{ON}$ **has a length of** $18$ **cm**.

### Example 2

In the triangle $\Delta ACB$, the line $\overline{CP}$ bisects $\angle ACB$. Suppose that $\overline{AC} = 36$ ft, $\overline{CB} = 42$ ft, and $\overline{AB} = 26$ ft, what is the length of line segment $\overline{PB}$?

__Solution__

Begin by constructing $\Delta ACB$ with the given components. Keep in mind that $\overline{CP}$ **divides the opposite side** $\overline{AB}$ **into two line segments**: $\overline{AP}$ and $\overline{PB}$. If $x$ represents the length of $\overline{PB}$, $\overline{AP}$ is equal to $(26 – x)$ ft.

Using the angle bisector theorem, **the ratio of** $\overline{AC}$** and** $\overline{AP}$ **is equal to** $\overline{CB}$ **and** $\overline{PB}$.

\begin{aligned}\dfrac{AC}{AP} &= \dfrac{CB}{PB}\\\dfrac{36}{26- x} &= \dfrac{42}{x}\end{aligned}

Apply cross-multiplication to simplify and solve the resulting equation. Find the length of $\overline{PB}$ by** finding the value of** $x$.

\begin{aligned}36x &= 42(26- x)\\36x &= 1092- 42x\\36x + 42x &= 1092\\78x &= 1092\\x&= 14\end{aligned}

Hence, **the length of** $\overline{PB}$ **is equal to** $14$** ft**.

*Practice Question*

1. In the triangle $\Delta LMN$ the line $\overline{MO}$ bisects $\angle LMO$. Suppose that $\overline{LM} = 20$ cm, $\overline{MN} = 81$ cm, and $\overline{LO} = 64$ cm, what is the length of line segment $\overline{ON}$?

A. $\overline{ON} = 45$ cm

B. $\overline{ON} = 64$ cm

C. $\overline{ON} = 72$ cm

D. $\overline{ON} = 81$ cm

2. In the triangle $\Delta ACB$, the line $\overline{CP}$ bisects $\angle ACB$. Suppose that $\overline{AC} = 38$ ft, $\overline{CB} = 57$ ft, and $\overline{AB} = 75$ ft, what is the length of line segment $\overline{PB}$?

A. $\overline{PB} = 38$ ft

B. $\overline{PB} = 45$ ft

C. $\overline{PB} = 51$ ft

D. $\overline{PB} = 57$ ft

3. The angle bisector $\overline{AD}$ divides the line segment $AC$ that forms the triangle $\Delta ACB$. Suppose that $\overline{AC} = 12$ m, $\overline{CB} = 37$ m, and $\overline{AB} = 14$ m, what is the length of line segment $\overline{CD}$?

A. $\overline{CD} = 18$ cm

B. $\overline{CD} = 21$ cm

C. $\overline{CD} = 24$ m

D. $\overline{CD} = 30$ cm

*Answer Key*

1. C

2. B

3. A