What if the cities of Verticville, Triopolis, and Angletown were joining their city budgets together to build a centrally located airport? There are freeways between the three cities and they want to have the freeway on the interior of these freeways. Where is the best location to put the airport so that they have to build the least amount of road? In the picture below, the blue lines are the proposed roads. After completing this Concept, you'll be able to use angle bisectors to help answer this question.

### Watch This

CK-12 Foundation: Chapter5AngleBisectorsA

James Sousa: Introduction to Angle Bisectors

James Sousa: Proof of the Angle Bisector Theorem

James Sousa: Proof of the Angle Bisector Theorem Converse

James Sousa: Solving For Unknown Values Using Angle Bisectors

### Guidance

Recall that an **angle bisector** cuts an angle exactly in half. Let’s analyze this figure.

\begin{align*}\overrightarrow{BD}\end{align*} is the angle bisector of \begin{align*}\angle ABC\end{align*}. Looking at point \begin{align*}D\end{align*}, if we were to draw \begin{align*}\overline{ED}\end{align*} and \begin{align*}\overline{DF}\end{align*}, we would find that they are equal. Recall that the shortest distance from a point to a line is the perpendicular length between them. \begin{align*}ED\end{align*} and \begin{align*}DF\end{align*} are the shortest lengths between \begin{align*}D\end{align*}, *which is on the angle bisector*, and each side of the angle.

**Angle Bisector Theorem:** If a point is on the bisector of an angle, then the point is equidistant from the sides of the angle.

In other words, if \begin{align*}\overleftrightarrow{BD}\end{align*} bisects \begin{align*}\angle ABC, \overrightarrow{BE} \bot \overline{ED}\end{align*}, and \begin{align*}\overrightarrow{BF} \bot \overline{DF}\end{align*}, then \begin{align*}ED = DF\end{align*}.

**Proof of the Angle Bisector Theorem:**

Given: \begin{align*}\overrightarrow{BD} \end{align*} bisects \begin{align*}\angle ABC, \overrightarrow{BA} \bot \overline{AD}\end{align*}, and \begin{align*}\overrightarrow{BC} \bot \overline{DC}\end{align*}

Prove: \begin{align*}\overline{AD} \cong \overline{DC}\end{align*}

Statement |
Reason |
---|---|

1. \begin{align*}\overrightarrow{BD}\end{align*} bisects \begin{align*}\angle ABC, \overrightarrow{BA} \bot \overline{AD}, \overrightarrow{BC} \bot \overline{DC}\end{align*} | Given |

2. \begin{align*}\angle ABD \cong \angle DBC\end{align*} | Definition of an angle bisector |

3. \begin{align*}\angle DAB\end{align*} and \begin{align*}\angle DCB\end{align*} are right angles | Definition of perpendicular lines |

4. \begin{align*}\angle DAB \cong \angle DCB\end{align*} | All right angles are congruent |

5. \begin{align*}\overline{BD} \cong \overline{BD}\end{align*} | Reflexive PoC |

6. \begin{align*}\triangle ABD \cong \triangle CBD\end{align*} | AAS |

7. \begin{align*}\overline{AD} \cong \overline{DC}\end{align*} | CPCTC |

The converse of this theorem is also true.

**Angle Bisector Theorem Converse:** If a point is in the interior of an angle and equidistant from the sides, then it lies on the bisector of the angle.

Because the Angle Bisector Theorem and its converse are both true we have a biconditional statement. We can put the two conditional statements together using if and only if. *A point is on the angle bisector of an angle if and only if it is equidistant from the sides of the triangle.* Like perpendicular bisectors, the point of concurrency for angle bisectors has interesting properties.

##### Investigation: Constructing Angle Bisectors in Triangles

Tools Needed: compass, ruler, pencil, paper

1. Draw a scalene triangle. Construct the angle bisector of each angle. Use Investigation 1-4 and #1 from the Review Queue to help you.

**Incenter:** The point of concurrency for the angle bisectors of a triangle.

2. Erase the arc marks and the angle bisectors after the incenter. Draw or construct the perpendicular lines to each side, through the incenter.

3. Erase the arc marks from #2 and the perpendicular lines beyond the sides of the triangle. Place the pointer of the compass on the incenter. Open the compass to intersect one of the three perpendicular lines drawn in #2. Draw a circle.

Notice that the circle touches all three sides of the triangle. We say that this circle is ** inscribed** in the triangle because it touches all three sides. The incenter is on all three angle bisectors, so

*the incenter is equidistant from all three sides of the triangle.*
**Concurrency of Angle Bisectors Theorem:** The angle bisectors of a triangle intersect in a point that is equidistant from the three sides of the triangle.

If \begin{align*}\overline{AG}, \overline{BG}\end{align*}, and \begin{align*}\overline{GC}\end{align*} are the angle bisectors of the angles in the triangle, then \begin{align*}EG = GF = GD\end{align*}.

In other words, \begin{align*}\overline{EG}, \overline{FG}\end{align*}, and \begin{align*}\overline{DG}\end{align*} are the radii of the inscribed circle.

#### Example A

Is \begin{align*}Y\end{align*} on the angle bisector of \begin{align*}\angle XWZ\end{align*}?

In order for \begin{align*}Y\end{align*} to be on the angle bisector \begin{align*}XY\end{align*} needs to be equal to \begin{align*}YZ\end{align*} and they both need to be perpendicular to the sides of the angle. From the markings we know \begin{align*}\overline{XY} \bot \overrightarrow{WX}\end{align*} and \begin{align*}\overline{ZY} \bot \overrightarrow{WZ}\end{align*}. Second, \begin{align*}XY = YZ = 6\end{align*}. From this we can conclude that \begin{align*}Y\end{align*} is on the angle bisector.

#### Example B

If \begin{align*}J, E\end{align*}, and \begin{align*}G\end{align*} are midpoints and \begin{align*}KA = AD = AH\end{align*} what are points \begin{align*}A\end{align*} and \begin{align*}B\end{align*} called?

\begin{align*}A\end{align*} is the incenter because \begin{align*}KA = AD = AH\end{align*}, which means that it is equidistant to the sides. \begin{align*}B\end{align*} is the circumcenter because \begin{align*}\overline{JB}, \overline{BE}\end{align*}, and \begin{align*}\overline{BG}\end{align*} are the perpendicular bisectors to the sides.

#### Example C

\begin{align*}\overrightarrow{AB}\end{align*} is the angle bisector of \begin{align*}\angle CAD\end{align*}. Solve for the missing variable.

\begin{align*}CB=BD\end{align*} by the Angle Bisector Theorem, so we can set up and solve an equation for \begin{align*}x\end{align*}.

\begin{align*} x+7 &=2(3x-4)\\ x+7 &=6x-8\\ 15 &=5x\\ x &=3 \end{align*}

Watch this video for help with the Examples above.

CK-12 Foundation: Chapter5AngleBisectorsB

#### Concept Problem Revisited

The airport needs to be equidistant to the three highways between the three cities. Therefore, the roads are all perpendicular to each side and congruent. The airport should be located at the incenter of the triangle.

### Vocabulary

An ** angle bisector** cuts an angle exactly in half.

**means the same distance from. A point is equidistant from two lines if it is the same distance from both lines. When we construct angle bisectors for the angles of a triangle, they meet in one point. This point is called the**

*Equidistant***of the triangle.**

*incenter*### Guided Practice

1. Is there enough information to determine if \begin{align*}\overrightarrow{A B}\end{align*} is the angle bisector of \begin{align*}\angle CAD\end{align*}? Why or why not?

2. \begin{align*}\overrightarrow{MO}\end{align*} is the angle bisector of \begin{align*}\angle LMN\end{align*}. Find the measure of \begin{align*}x\end{align*}.

3. A \begin{align*} 100^\circ\end{align*} angle is bisected. What are the measures of the resulting angles?

**Answers:**

1. No because \begin{align*}B\end{align*} is not necessarily equidistant from \begin{align*}\overline{AC}\end{align*} and \begin{align*}\overline{AD}\end{align*}. We do not know if the angles in the diagram are right angles.

2. \begin{align*}LO = ON\end{align*} by the Angle Bisector Theorem.

\begin{align*}4x - 5 &= 23\\ 4x &= 28\\ x &=7\end{align*}

3. We know that to bisect means to cut in half, so each of the resulting angles will be half of \begin{align*}100\end{align*}. The measure of each resulting angle is \begin{align*}50^\circ\end{align*}.

### Interactive Practice

### Practice

For questions 1-6, \begin{align*}\overrightarrow{AB}\end{align*} is the angle bisector of \begin{align*}\angle CAD\end{align*}. Solve for the missing variable.

Is there enough information to determine if \begin{align*}\overrightarrow{AB}\end{align*} is the angle bisector of \begin{align*}\angle CAD\end{align*}? Why or why not?

- Fill in the blanks in the Angle Bisector Theorem Converse.

Given: \begin{align*}\overline{AD} \cong \overline{DC}\end{align*}, such that \begin{align*}AD\end{align*} and \begin{align*}DC\end{align*} are the shortest distances to \begin{align*}\overrightarrow{BA}\end{align*} and \begin{align*}\overrightarrow{BC}\end{align*}

Prove: \begin{align*}\overrightarrow{BD}\end{align*} bisects \begin{align*}\angle ABC\end{align*}

Statement |
Reason |
---|---|

1. | |

2. | The shortest distance from a point to a line is perpendicular. |

3. \begin{align*}\angle DAB\end{align*} and \begin{align*}\angle DCB\end{align*} are right angles | |

4. \begin{align*}\angle DAB \cong \angle DCB\end{align*} | |

5. \begin{align*}\overline{BD} \cong \overline{BD}\end{align*} | |

6. \begin{align*}\triangle ABD \cong \triangle CBD\end{align*} | |

7. | CPCTC |

8. \begin{align*}\overrightarrow{BD}\end{align*} bisects \begin{align*}\angle ABC\end{align*} |

Determine if the following descriptions refer to the incenter or circumcenter of the triangle.

- A lighthouse on a triangular island is equidistant to the three coastlines.
- A hospital is equidistant to three cities.
- A circular walking path passes through three historical landmarks.
- A circular walking path connects three other straight paths.

*Multi- Step Problem*

- Draw \begin{align*}\angle ABC\end{align*} through \begin{align*}A(1, 3), B(3, -1)\end{align*} and \begin{align*}C(7, 1)\end{align*}.
- Use slopes to show that \begin{align*}\angle ABC\end{align*} is a right angle.
- Use the distance formula to find \begin{align*}AB\end{align*} and \begin{align*}BC\end{align*}.
- Construct a line perpendicular to \begin{align*}AB\end{align*} through \begin{align*}A\end{align*}.
- Construct a line perpendicular to \begin{align*}BC\end{align*} through \begin{align*}C\end{align*}.
- These lines intersect in the interior of \begin{align*}\angle ABC\end{align*}. Label this point \begin{align*}D\end{align*} and draw \begin{align*}\overrightarrow{BD}\end{align*}.
- Is \begin{align*}\overrightarrow{BD}\end{align*} the angle bisector of \begin{align*}\angle ABC\end{align*}? Justify your answer.