5.2: Perpendicular Bisectors and Angle Bisectors in Triangles
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
 Construct the perpendicular bisector of a 3 inch line. Use Investigation 14 from Chapter 1 to help you.
 Construct the angle bisector of an
80∘ angle (Investigation 15).  Find the value of
x .  Find the value of
x andy . Ism the perpendicular bisector ofAB ? 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 51: 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:
What do you notice about these distances?
From the investigation, you should notice that
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
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
Proof of the Perpendicular Bisector Theorem Converse
Given:
Prove:
Example 1: If
Solution: By the Perpendicular Bisector Theorem,
Example 2: Algebra Connection Find
Solution:
Example 3:
a) Which line segments are equal?
b) Find
c) Is
Solution:
a)
b)
c) Yes,
Perpendicular Bisectors and Triangles
Let’s investigate what happens when we construct perpendicular lines for the sides of a triangle.
Investigation 52: 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 53: 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,
2. Recall the patty paper construct of the perpendicular bisector above (Investigation 52). Using this idea, fold a perpendicular line to
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 54: 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 52 to find the perpendicular bisector of at least two sides.
Review Questions
 Questions 13 are similar to Investigation 52.
 Questions 46 are similar to Investigation 54.
 Questions 715 are similar to Examples 13.
 Questions 1624 are similar to Examples 4 and 5.
 Question 2528 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 52.
 Construct equilateral triangle \begin{align*}\triangle ABC\end{align*} (Investigation 46). 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 54. Construct the inscribed circle.
 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 712, find the value of \begin{align*}x\end{align*}. \begin{align*}m\end{align*} is the perpendicular bisector of \begin{align*}AB\end{align*}.

\begin{align*}m\end{align*} is the perpendicular bisector of \begin{align*}\overline{AB}\end{align*}.
 List all the congruent segments.
 Is \begin{align*}C\end{align*} on \begin{align*}\overline{AB}\end{align*}? Why or why not?
 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 611, \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 2528, 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*}.
 Find the slope of \begin{align*}AB\end{align*}.
 Find the midpoint of \begin{align*}AB\end{align*}.
 What is the slope of the perpendicular line to \begin{align*}AB\end{align*}?
 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*}.
 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*} 
 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*} 
Review Queue Answers
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*}6x13=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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