5.2: Perpendicular Bisectors
What if an archeologist found three bones in Cairo, Egypt buried 4 meters, 7 meters, and 9 meters apart (to form a triangle)? The likelihood that more bones are in this area is very high. The archeologist wants to dig in an appropriate circle around these bones. If these bones are on the edge of the digging circle, where is the center of the circle? Can you determine how far apart each bone is from the center of the circle? What is this length? After completing this Concept, you'll be able to answer questions like these.
Watch This
CK12 Foundation: Chapter5PerpendicularBisectorsA
James Sousa: Constructing Perpendicular Bisectors
James Sousa: Proof of the Perpendicular Bisector Theorem
James Sousa: Proof of the Perpendicular Bisector Theorem Converse
James Sousa: Determining Values Using Perpendicular Bisectors
Guidance
Recall that a perpendicular bisector intersects a line segment at its midpoint and is perpendicular. Let’s analyze this figure.
\begin{align*}\overleftrightarrow{CD}\end{align*} is the perpendicular bisector of \begin{align*}\overline{AB}\end{align*}. If we were to draw in \begin{align*}\overline{AC}\end{align*} and \begin{align*}\overline{CB}\end{align*}, we would find that they are equal. Therefore, any point on the perpendicular bisector of a segment is the same distance from each endpoint.
Perpendicular Bisector Theorem: If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.
In addition to the Perpendicular Bisector Theorem, we also know that its converse is 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.
Proof of the Perpendicular Bisector Theorem Converse:
Given: \begin{align*}\overline{AC} \cong \overline{CB}\end{align*}
Prove: \begin{align*}\overleftrightarrow{CD}\end{align*} is the perpendicular bisector of \begin{align*}\overline{AB}\end{align*}
Statement  Reason 

1. \begin{align*}\overline{AC} \cong \overline{CB}\end{align*}  Given 
2. \begin{align*}\triangle ACB\end{align*} is an isosceles triangle  Definition of an isosceles triangle 
3. \begin{align*}\angle A \cong \angle B\end{align*}  Isosceles Triangle Theorem 
4. Draw point \begin{align*}D\end{align*}, such that \begin{align*}D\end{align*} is the midpoint of \begin{align*}\overline{AB}\end{align*}.  Every line segment has exactly one midpoint 
5. \begin{align*}\overline{AD} \cong \overline{DB}\end{align*}  Definition of a midpoint 
6. \begin{align*}\triangle ACD \cong \triangle BCD\end{align*}  SAS 
7. \begin{align*}\angle CDA \cong \angle CDB\end{align*}  CPCTC 
8. \begin{align*}m \angle CDA=m \angle CDB=90^\circ\end{align*}  Congruent Supplements Theorem 
9. \begin{align*}\overleftrightarrow{CD} \bot \overline{AB}\end{align*}  Definition of perpendicular lines 
10. \begin{align*}\overleftrightarrow{CD}\end{align*} is the perpendicular bisector of \begin{align*}\overline{AB}\end{align*}  Definition of perpendicular bisector 
Two lines intersect at a point. If more than two lines intersect at the same point, it is called a point of concurrency.
Investigation: Constructing the Perpendicular Bisectors of the Sides of a Triangle
Tools Needed: paper, pencil, compass, ruler
1. Draw a scalene triangle.
2. Construct the perpendicular bisector for all three sides.
The three perpendicular bisectors all intersect at the same point, called the circumcenter.
Circumcenter: The point of concurrency for the perpendicular bisectors of the sides of a triangle.
3. Erase the arc marks to leave only the perpendicular bisectors. Put the pointer of your compass on the circumcenter. Open the compass so that the pencil is on one of the vertices. Draw a circle. What happens?
The circumcenter is the center of a circle that passes through all the vertices of the triangle. We say that this circle circumscribes the triangle. This means that the circumcenter is equidistant to the vertices.
Concurrency of Perpendicular Bisectors Theorem: The perpendicular bisectors of the sides of a triangle intersect in a point that is equidistant from the vertices.
If \begin{align*}\overline{PC}, \overline{QC}\end{align*}, and \begin{align*}\overline{RC}\end{align*} are perpendicular bisectors, then \begin{align*}LC=MC=OC\end{align*}.
Example A
Find \begin{align*}x\end{align*} and the length of each segment.
From the markings, we know that \begin{align*}\overleftrightarrow{WX}\end{align*} is the perpendicular bisector of \begin{align*}\overline{XY}\end{align*}. Therefore, we can use the Perpendicular Bisector Theorem to conclude that \begin{align*}WZ = WY\end{align*}. Write an equation.
\begin{align*}2x+11 &= 4x5\\ 16 &= 2x\\ 8 &= x\end{align*}
To find the length of \begin{align*}WZ\end{align*} and \begin{align*}WY\end{align*}, substitute 8 into either expression, \begin{align*}2(8)+11=16+11=27\end{align*}.
Example B
\begin{align*}\overleftrightarrow{OQ}\end{align*} is the perpendicular bisector of \begin{align*}\overline{MP}\end{align*}.
a) Which segments are equal?
b) Find \begin{align*}x\end{align*}.
c) Is \begin{align*}L\end{align*} on \begin{align*}\overleftrightarrow{OQ}\end{align*}? How do you know?
Answer:
a) \begin{align*}ML = LP\end{align*} because they are both 15.
\begin{align*}MO = OP\end{align*} because \begin{align*}O\end{align*} is the midpoint of \begin{align*}\overline{MP}\end{align*}
\begin{align*}MQ = QP\end{align*} because \begin{align*}Q\end{align*} is on the perpendicular bisector of \begin{align*}\overline{MP}\end{align*}.
b) \begin{align*}4x+3 &= 11 \\ 4x &= 8\\ x &= 2\end{align*}
c) Yes, \begin{align*}L\end{align*} is on \begin{align*}\overleftrightarrow{OQ}\end{align*} because \begin{align*}ML = LP\end{align*} (Perpendicular Bisector Theorem Converse).
Example C
For further exploration, try the following:
 Cut out an acute triangle from a sheet of paper.
 Fold the triangle over one side so that the side is folded in half. Crease.
 Repeat for the other two sides. What do you notice?
The folds (blue dashed lines)are the perpendicular bisectors and cross at the circumcenter.
Watch this video for help with the Examples above.
CK12 Foundation: Chapter5PerpendicularBisectorsB
Concept Problem Revisited
The center of the circle will be the circumcenter of the triangle formed by the three bones. Construct the perpendicular bisector of at least two sides to find the circumcenter. After locating the circumcenter, the archeologist can measure the distance from each bone to it, which would be the radius of the circle. This length is approximately 4.7 meters.
Vocabulary
Perpendicular lines are lines that meet at right (\begin{align*}90^\circ\end{align*}) angles. A midpoint is the point on a segment that divides the segment into two equal parts. A perpendicular bisector is a line that intersects a line segment at its midpoint and is perpendicular to that line segment. When we construct perpendicular bisectors for the sides of a triangle, they meet in one point. This point is called the circumcenter of the triangle.
Guided Practice
1. Find the value of \begin{align*}x\end{align*}. \begin{align*}m\end{align*} is the perpendicular bisector of \begin{align*}AB\end{align*}.
2. 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.
Answers:
1. By the Perpendicular Bisector Theorem, both segments are equal. Set up and solve an equation.
\begin{align*}x+6 &=22\\ x &=16\end{align*}
2. \begin{align*}\overleftrightarrow{S T}\end{align*} is not necessarily the perpendicular bisector of \begin{align*}\overline{XY}\end{align*} because not enough information is given in the diagram. There is no way to know from the diagram if \begin{align*}\overleftrightarrow{S T}\end{align*} will extend to make a right angle with \begin{align*}\overline{XY}\end{align*}.
Practice

\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 Question 2, determine if \begin{align*}\overleftrightarrow{ST}\end{align*} is the perpendicular bisector of \begin{align*}\overline{XY}\end{align*}. Explain why or why not.
For Questions 37, 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*}.
 Find the equation of the perpendicular bisector of \begin{align*}AB\end{align*}.
 Find \begin{align*}AB\end{align*}. Simplify the radical, if needed.
 Plot \begin{align*}A, B\end{align*}, and the perpendicular bisector. Label it \begin{align*}m\end{align*}. How could you find a point \begin{align*}C\end{align*} on \begin{align*}m\end{align*}, such that \begin{align*}C\end{align*} would be the third vertex of equilateral triangle \begin{align*}\triangle ABC\end{align*}? You do not have to find the coordinates, just describe how you would do it.
For Questions 812, consider \begin{align*}\triangle ABC\end{align*} with vertices \begin{align*}A(7, 6), B(7, 2)\end{align*} and \begin{align*}C(0, 5)\end{align*}. Plot this triangle on graph paper.
 Find the midpoint and slope of \begin{align*}\overline{AB}\end{align*} and use them to draw the perpendicular bisector of \begin{align*}\overline{AB}\end{align*}. You do not need to write the equation.
 Find the midpoint and slope of \begin{align*}\overline{BC}\end{align*} and use them to draw the perpendicular bisector of \begin{align*}\overline{BC}\end{align*}. You do not need to write the equation.
 Find the midpoint and slope of \begin{align*}\overline{AC}\end{align*} and use them to draw the perpendicular bisector of \begin{align*}\overline{AC}\end{align*}. You do not need to write the equation.
 Are the three lines concurrent? What are the coordinates of their point of intersection (what is the circumcenter of the triangle)?
 Use your compass to draw the circumscribed circle about the triangle with your point found in question 11 as the center of your circle.
 Fill in the blanks: There is exactly _________ circle which contains any __________ points.
 Fill in the blanks of the proof of the Perpendicular Bisector Theorem.
Given: \begin{align*}\overleftrightarrow{CD}\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*} 
 Write a two column proof.
Given: \begin{align*}\triangle ABC\end{align*} is a right isosceles triangle and \begin{align*}\overline{BD}\end{align*} is the \begin{align*}\bot\end{align*} bisector of \begin{align*}\overline{AC}\end{align*}
Prove: \begin{align*}\triangle ABD\end{align*} and \begin{align*}\triangle CBD\end{align*} are congruent.
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Term  Definition 

circumcenter  The circumcenter is the point of intersection of the perpendicular bisectors of the sides in a triangle. 
perpendicular bisector  A perpendicular bisector of a line segment passes through the midpoint of the line segment and intersects the line segment at . 
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. 
Image Attributions
Here you'll learn what a perpendicular bisector is and the Perpendicular Bisector Theorem, which states that if a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.