JUMP TO TOPIC

**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.

**Thanks to triangle theorems like this,**

*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.***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, $\angle 2 = \angle 4$.

**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.

**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} |

**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**.