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# 5.2: Perpendicular Bisectors and Angle Bisectors in Triangles

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Apply the Perpendicular Bisector Theorem and its converse.
• Apply the Angle Bisector Theorem and its converse.
• Analyze properties of perpendicular bisectors and angle bisectors.

## Review Queue

1. Construct the perpendicular bisector of a 3 inch line. Use Investigation 1-4 from Chapter 1 to help you.
2. Construct the angle bisector of an 80\begin{align*}80^\circ\end{align*} angle (Investigation 1-5).
3. Find the value of x\begin{align*}x\end{align*}.
4. Find the value of x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*}. Is m\begin{align*}m\end{align*} the perpendicular bisector of AB\begin{align*}AB\end{align*}? How do you know?

Know What? An archeologist has found three bones in Cairo, Egypt. The bones are 4 meters apart, 7 meters apart and 9 meters apart (to form a triangle). The likelihood that more bones are in this area is very high. If these bones are on the edge of the digging circle, where is the center of the circle?

## Perpendicular Bisectors

In Chapter 1, you learned that a perpendicular bisector intersects a line segment at its midpoint and is perpendicular. Let’s analyze your construction from #1.

Investigation 5-1: Properties of Perpendicular Bisectors

Tools Needed: #1 from the Review Queue, ruler, pencil

1. Look at your construction (#1 from the Review Queue).

Draw three points on the perpendicular bisector, two above the line and one below it. Label all the points like the picture on the right.

2. Measure the following distances with a ruler: AD, DB, AC, CB, AE\begin{align*}AD, \ DB, \ AC, \ CB, \ AE\end{align*}, and EB\begin{align*}EB\end{align*}. Record them on your paper.

What do you notice about these distances?

From the investigation, you should notice that AD=DB, AC=CB\begin{align*}AD = DB, \ AC = CB\end{align*}, and AE=EB\begin{align*}AE = EB\end{align*}. This means that C, D\begin{align*}C, \ D\end{align*}, and E\begin{align*}E\end{align*} are equidistant from A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*}.

Perpendicular Bisector Theorem: If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.

If CDAB¯¯¯¯¯¯¯¯\begin{align*}\overleftrightarrow{CD} \perp \overline{AB}\end{align*} and AD=DB\begin{align*}AD = DB\end{align*}, then AC=CB\begin{align*}AC = CB\end{align*}.

The proof of the Perpendicular Bisector Theorem is in the exercises for this section. In addition to the Perpendicular Bisector Theorem, the converse is also true.

Perpendicular Bisector Theorem Converse: If a point is equidistant from the endpoints of a segment, then the point is on the perpendicular bisector of the segment.

Using the picture above: If AC=CB\begin{align*}AC = CB\end{align*}, then CDAB¯¯¯¯¯¯¯¯\begin{align*}\overleftrightarrow{CD} \perp \overline{AB}\end{align*} and AD=DB\begin{align*}AD = DB\end{align*}.

Proof of the Perpendicular Bisector Theorem Converse

Given: AC¯¯¯¯¯¯¯¯CB¯¯¯¯¯¯¯¯\begin{align*}\overline{AC} \cong \overline{CB}\end{align*}

Prove: CD\begin{align*}\overleftrightarrow{C D}\end{align*} is the perpendicular bisector of AB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB}\end{align*}

Example 1: If MO\begin{align*}\overleftrightarrow{MO}\end{align*} is the perpendicular bisector of LN¯¯¯¯¯¯¯¯\begin{align*}\overline{LN}\end{align*} and LO=8\begin{align*}LO = 8\end{align*}, what is ON\begin{align*}ON\end{align*}?

Solution: By the Perpendicular Bisector Theorem, LO=ON\begin{align*}LO = ON\end{align*}. So, ON=8\begin{align*}ON = 8\end{align*}.

Example 2: Algebra Connection Find x\begin{align*}x\end{align*} and the length of each segment.

Solution: WX\begin{align*}\overleftrightarrow{WX}\end{align*} is the perpendicular bisector of XY¯¯¯¯¯¯¯¯\begin{align*}\overline{XY}\end{align*} and from the Perpendicular Bisector Theorem WZ=WY\begin{align*}WZ = WY\end{align*}.

2x+11168=4x5=2x=x\begin{align*}2x + 11 &= 4x - 5\\ 16 &= 2x\\ 8 &= x\end{align*}

WZ=WY=2(8)+11=16+11=27\begin{align*}WZ = WY = 2(8) + 11 = 16 + 11 = 27\end{align*}.

Example 3: OQ\begin{align*}\overleftrightarrow{OQ}\end{align*} is the perpendicular bisector of MP¯¯¯¯¯¯¯¯¯\begin{align*}\overline{MP}\end{align*}.

a) Which line segments are equal?

b) Find x\begin{align*}x\end{align*}.

c) Is L\begin{align*}L\end{align*} on OQ\begin{align*}\overleftrightarrow{OQ}\end{align*}? How do you know?

Solution:

a) ML=LP, MO=OP\begin{align*}ML = LP, \ MO = OP\end{align*}, and MQ=QP\begin{align*}MQ = QP\end{align*}.

b) 4x+3=114x=8x=2\begin{align*}4x + 3 = 11\!\\ 4x = 8\!\\ x = 2\end{align*}

c) Yes, L\begin{align*}L\end{align*} is on OQ\begin{align*}\overleftrightarrow{OQ}\end{align*} because ML=LP\begin{align*}ML = LP\end{align*} (the Perpendicular Bisector Theorem Converse).

## Perpendicular Bisectors and Triangles

Let’s investigate what happens when we construct perpendicular lines for the sides of a triangle.

Investigation 5-2: Constructing the Perpendicular Bisectors of the Sides of a Triangle

Tools Needed: patty paper, pencil, ruler, compass

1. Draw a scalene triangle on your patty paper.

2. Fold one vertex over to meet one of the other vertices. Make sure one side perfectly overlaps itself. Crease and open.

3. Repeat this process for the other two sides. Each crease is the perpendicular bisector of a side. Your paper should look like this:

The creases, or perpendicular bisectors, intersect at the same point.

4. This point has an additional property. Put the pointer of your compass on this point of intersection. Open the compass so that the pencil is on one of the vertices. Draw a circle.

The circle you drew passes through all the vertices of the triangle. We say that this circle circumscribes the triangle or that the triangle is inscribed in the circle.

## Angle Bisectors

In Chapter 1, you learned that an angle bisector cuts an angle exactly in half.

Investigation 5-3: Properties of an Angle Bisector

Tools Needed: #2 from your Review Queue, protractor, ruler, pencil

1. Look at #2 from the Review Queue. Label your angle like the one to the right. Place two points, D\begin{align*}D\end{align*} and E\begin{align*}E\end{align*} on the angle bisector.

2. Recall the patty paper construct of the perpendicular bisector above (Investigation 5-2). Using this idea, fold a perpendicular line to BC\begin{align*}\overrightarrow{BC}\end{align*} through D\begin{align*}D\end{align*}. Repeat with \begin{align*}D\end{align*} and \begin{align*}\overrightarrow{BA}\end{align*}. Label the intersections \begin{align*}F\end{align*} and \begin{align*}G\end{align*}.

3. Measure \begin{align*}FD\end{align*} and \begin{align*}DG\end{align*}. What do you notice?

4. Repeat #2 and #3 with point \begin{align*}E\end{align*}. Do you have the same conclusion?

For #3, you should find that \begin{align*}FD = DG\end{align*} and the same thing happens with \begin{align*}E\end{align*}.

Recall from Chapter 3 that the shortest distance from a point to a line is the perpendicular length between them. \begin{align*}FD = DG\end{align*} and are the shortest lengths from \begin{align*}D\end{align*} to 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*}\overrightarrow{BD}\end{align*} bisects \begin{align*}\angle ABC, \ \overrightarrow{BA} \perp \overline{FD}\end{align*}, and, \begin{align*}\overrightarrow{BC} \perp \overline{DG}\end{align*} then \begin{align*}FD = DG\end{align*}.

Proof of the Angle Bisector Theorem

Given: \begin{align*}\overrightarrow{BD}\end{align*} bisects \begin{align*} \angle ABC, \ \overrightarrow{BA} \perp \overline{AD}\end{align*}, and \begin{align*}\overrightarrow{BC} \perp \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} \perp \overline{AD}, \ \overrightarrow{BC} \perp \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. The proof is in the review questions.

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.

Example 4: Is \begin{align*}Y\end{align*} on the angle bisector of \begin{align*}\angle XWZ\end{align*}?

Solution: If \begin{align*}Y\end{align*} is on the angle bisector, then \begin{align*}XY = YZ\end{align*} and they need to be perpendicular to the sides of the angle. From the markings we know \begin{align*}\overline{XY} \perp \overrightarrow{WX}\end{align*} and \begin{align*}\overline{ZY} \perp \overrightarrow{WZ}\end{align*}. Second, \begin{align*}XY = YZ = 6\end{align*}. So, yes, \begin{align*}Y\end{align*} is on the angle bisector of \begin{align*}\angle XWZ\end{align*}.

Example 5: Algebra Connection \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*}.

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

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

## Angle Bisectors in a Triangle

Let’s use patty paper to construct the angle bisector of every angle in a triangle.

Investigation 5-4: Constructing Angle Bisectors in Triangles

Tools Needed: patty paper, ruler, pencil, compass

1. Draw a scalene triangle on your patty paper.

2. Fold the patty paper so that two sides of the triangle perfectly overlap and the fold passes through the vertex between these sides. Crease and open.

3. Repeat Step 2 for the other two angles. Your paper should look like:

The creases, or angle bisectors, intersect at the same point.

4. This point has an additional property. Place the pointer of the compass on this point. Open the compass “straight down” so that the pencil touches one side of the triangle (the pink line in the picture to the right). 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.

Know What? Revisited If the bones are the vertices of a triangle, then the center of the circle will be the intersection of the perpendicular bisectors. Use Investigation 5-2 to find the perpendicular bisector of at least two sides.

## Review Questions

• Questions 1-3 are similar to Investigation 5-2.
• Questions 4-6 are similar to Investigation 5-4.
• Questions 7-15 are similar to Examples 1-3.
• Questions 16-24 are similar to Examples 4 and 5.
• Question 25-28 are a review of perpendicular lines.
• Questions 29 and 30 are similar to the two proofs in this section.

Construction Find the point of intersection of the perpendicular bisectors by tracing each triangle onto a piece of paper (or patty paper) and using Investigation 5-2.

1. Construct equilateral triangle \begin{align*}\triangle ABC\end{align*} (Investigation 4-6). Construct the perpendicular bisectors of the sides of the triangle. Connect the point of intersection to each vertex. Your original triangle is now divided into six triangles. What can you conclude about the six triangles?

Construction Find the point of intersection of the angle bisectors by tracing each triangle onto a piece of paper (or patty paper) and using Investigation 5-4. Construct the inscribed circle.

1. Refer back to #3, if you were to construct the angle bisectors of an equilateral triangle, what do you think would happen? Would the result of #3 be any different?

Algebra Connection For questions 7-12, find the value of \begin{align*}x\end{align*}. \begin{align*}m\end{align*} is the perpendicular bisector of \begin{align*}AB\end{align*}.

1. \begin{align*}m\end{align*} is the perpendicular bisector of \begin{align*}\overline{AB}\end{align*}.
1. List all the congruent segments.
2. Is \begin{align*}C\end{align*} on \begin{align*}\overline{AB}\end{align*}? Why or why not?
3. Is \begin{align*}D\end{align*} on \begin{align*}\overline{AB}\end{align*}? Why or why not?

For Questions 14 and 15, determine if \begin{align*}\overleftrightarrow{S T}\end{align*} is the perpendicular bisector of \begin{align*}\overline{XY}\end{align*}. Explain why or why not.

For questions 6-11, \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{A B}\end{align*} is the angle bisector of \begin{align*}\angle CAD\end{align*}? Why or why not?

For Questions 25-28, consider line segment \begin{align*}\overline{AB}\end{align*} with endpoints \begin{align*}A(2, 1)\end{align*} and \begin{align*}B(6, 3)\end{align*}.

1. Find the slope of \begin{align*}AB\end{align*}.
2. Find the midpoint of \begin{align*}AB\end{align*}.
3. What is the slope of the perpendicular line to \begin{align*}AB\end{align*}?
4. Find the equation of the line with the slope from #27 and through the midpoint from #26. This is the perpendicular bisector of \begin{align*}\overline{AB}\end{align*}.
5. Fill in the blanks of the proof of the Perpendicular Bisector Theorem. Given: \begin{align*}\overleftrightarrow{C D}\end{align*} is the perpendicular bisector of \begin{align*}\overline{AB}\end{align*} Prove: \begin{align*}\overline{AC} \cong \overline{CB}\end{align*}
Statement Reason
1.
2. \begin{align*}D\end{align*} is the midpoint of \begin{align*}\overline{AB}\end{align*}
3. Definition of a midpoint
4. \begin{align*}\angle CDA\end{align*} and \begin{align*}\angle CDB\end{align*} are right angles
5. \begin{align*}\angle CDA \cong \angle CDB\end{align*}
6. Reflexive PoC
7. \begin{align*}\triangle CDA \cong \triangle CDB\end{align*}
8. \begin{align*}\overline{AC} \cong \overline{CB}\end{align*}
1. 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{B D}\end{align*} bisects \begin{align*}\angle ABC\end{align*}

1.

2.

3. \begin{align*}2x+3=27\end{align*}

(a) \begin{align*}2x=24\!\\ {\;} \qquad x=12\end{align*}

(b) \begin{align*}5x+11=26\!\\ {\;}\qquad \quad \ \ 5x=15\!\\ {\;}\qquad \qquad \ x=3\end{align*}

4. \begin{align*}6x-13=2x+11 \qquad \qquad 3y+21=90^\circ \!\\ {\;}\qquad \quad 4x=24 \qquad \qquad \qquad \qquad \ 3y=69^\circ \!\\ {\;}\qquad \quad \ x=6 \qquad \qquad \qquad \qquad \quad \ y=23^\circ\end{align*}

Yes, \begin{align*}m\end{align*} is the perpendicular bisector of \begin{align*}AB\end{align*} because it is perpendicular to \begin{align*}AB\end{align*} and passes through the midpoint.

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