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# 5.2: Perpendicular Bisectors in Triangles

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Understand points of concurrency.
• Apply the Perpendicular Bisector Theorem and its converse to triangles.
• Understand concurrency for perpendicular bisectors.

## Review Queue

1. Construct the perpendicular bisector of a 3 inch line. Use Investigation 1-3 from Chapter 1 to help you.
2. Find the value of x\begin{align*}x\end{align*}.
3. 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. 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?

## Perpendicular Bisectors

In Chapter 1, you learned that a perpendicular bisector intersects a line segment at its midpoint and is perpendicular. In #1 in the Review Queue above, you constructed a perpendicular bisector of a 3 inch segment. Let’s analyze this figure.

CD\begin{align*}\overleftrightarrow{CD}\end{align*} is the perpendicular bisector of AB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB}\end{align*}. If we were to draw in AC¯¯¯¯¯¯¯¯\begin{align*}\overline{AC}\end{align*} and CB¯¯¯¯¯¯¯¯\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.

The proof of the Perpendicular Bisector Theorem is in the exercises for this section. 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: AC¯¯¯¯¯¯¯¯CB¯¯¯¯¯¯¯¯\begin{align*}\overline{AC} \cong \overline{CB}\end{align*}

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

Statement Reason
1. AC¯¯¯¯¯¯¯¯CB¯¯¯¯¯¯¯¯\begin{align*}\overline{AC} \cong \overline{CB}\end{align*} Given
2. ACB\begin{align*}\triangle ACB\end{align*} is an isosceles triangle Definition of an isosceles triangle
3. AB\begin{align*}\angle A \cong \angle B\end{align*} Isosceles Triangle Theorem
4. Draw point D\begin{align*}D\end{align*}, such that D\begin{align*}D\end{align*} is the midpoint of AB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB}\end{align*}. Every line segment has exactly one midpoint
5. AD¯¯¯¯¯¯¯¯DB¯¯¯¯¯¯¯¯\begin{align*}\overline{AD} \cong \overline{DB}\end{align*} Definition of a midpoint
6. ACDBCD\begin{align*}\triangle ACD \cong \triangle BCD\end{align*} SAS
7. CDACDB\begin{align*}\angle CDA \cong \angle CDB\end{align*} CPCTC
8. mCDA=mCDB=90\begin{align*}m \angle CDA=m \angle CDB=90^\circ\end{align*} Congruent Supplements Theorem
9. CDAB¯¯¯¯¯¯¯¯\begin{align*}\overleftrightarrow{CD} \bot \overline{AB}\end{align*} Definition of perpendicular lines
10. CD\begin{align*}\overleftrightarrow{CD}\end{align*} is the perpendicular bisector of AB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB}\end{align*} Definition of perpendicular bisector

Let’s use the Perpendicular Bisector Theorem and its converse in a few examples.

Example 1: Algebra Connection Find x\begin{align*}x\end{align*} and the length of each segment.

Solution: From the markings, we know that WX\begin{align*}\overleftrightarrow{WX}\end{align*} is the perpendicular bisector of XY¯¯¯¯¯¯¯¯\begin{align*}\overline{XY}\end{align*}. Therefore, we can use the Perpendicular Bisector Theorem to conclude that WZ=WY\begin{align*}WZ = WY\end{align*}. Write an equation.

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

To find the length of WZ\begin{align*}WZ\end{align*} and WY\begin{align*}WY\end{align*}, substitute 8 into either expression, 2(8)+11=16+11=27\begin{align*}2(8)+11=16+11=27\end{align*}.

Example 2: OQ\begin{align*}\overleftrightarrow{OQ}\end{align*} is the perpendicular bisector of MP¯¯¯¯¯¯¯¯¯\begin{align*}\overline{MP}\end{align*}.

a) Which 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\begin{align*}ML = LP\end{align*} because they are both 15.

MO=OP\begin{align*}MO = OP\end{align*} because O\begin{align*}O\end{align*} is the midpoint of MP¯¯¯¯¯¯¯¯¯\begin{align*}\overline{MP}\end{align*}

MQ=QP\begin{align*}MQ = QP\end{align*} because Q\begin{align*}Q\end{align*} is on the perpendicular bisector of MP¯¯¯¯¯¯¯¯¯\begin{align*}\overline{MP}\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*} (Perpendicular Bisector Theorem Converse).

## Perpendicular Bisectors and Triangles

Two lines intersect at a point. If more than two lines intersect at the same point, it is called a point of concurrency.

Point of Concurrency: When three or more lines intersect at the same point.

Investigation 5-1: 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 (Investigation 1-3) 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 PC¯¯¯¯¯¯¯¯,QC¯¯¯¯¯¯¯¯\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 3: For further exploration, try the following:

1. Cut out an acute triangle from a sheet of paper.
2. Fold the triangle over one side so that the side is folded in half. Crease.
3. Repeat for the other two sides. What do you notice?

Solution: The folds (blue dashed lines)are the perpendicular bisectors and cross at the circumcenter.

Know What? 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.

## Review Questions

Construction Construct the circumcenter for the following triangles by tracing each triangle onto a piece of paper and using Investigation 5-1.

1. Can you use the method in Example 3 to locate the circumcenter for these three triangles?
2. Based on your constructions in 1-3, state a conjecture about the relationship between a triangle and the location of its circumcenter.
3. Construct equilateral triangle \begin{align*}\triangle ABC\end{align*} (Investigation 4-6). Construct the perpendicular bisectors of the sides of the triangle and label the circumcenter \begin{align*}X\end{align*}. Connect the circumcenter to each vertex. Your original triangle is now divided into six triangles. What can you conclude about the six triangles? Why?

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{ST}\end{align*} is the perpendicular bisector of \begin{align*}\overline{XY}\end{align*}. Explain why or why not.

For Questions 16-20, 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. Find the equation of the perpendicular bisector of \begin{align*}AB\end{align*}.
4. Find \begin{align*}AB\end{align*}. Simplify the radical, if needed.
5. 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 21-25, 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.

1. 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.
2. 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.
3. 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.
4. Are the three lines concurrent? What are the coordinates of their point of intersection (what is the circumcenter of the triangle)?
5. Use your compass to draw the circumscribed circle about the triangle with your point found in question 24 as the center of your circle.
6. Repeat questions 21-25 with \begin{align*}\triangle LMO\end{align*} where \begin{align*}L(2, 9), M(3, 0)\end{align*} and \begin{align*}O(-7, 0)\end{align*}.
7. Repeat questions 21-25 with \begin{align*}\triangle REX\end{align*} where \begin{align*}R(4, 2), E(6, 0)\end{align*} and \begin{align*}X(0, 0)\end{align*}.
8. Can you explain why the perpendicular bisectors of the sides of a triangle would all pass through the center of the circle containing the vertices of the triangle? Think about the definition of a circle: The set of all point equidistant from a given center.
9. Fill in the blanks: There is exactly _________ circle which contains any __________ points.
10. 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*}
1. 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.
2. Write a paragraph explaining why the two smaller triangles in question 31 are also isosceles right triangles.

1. Reference Investigation 1-3.
1. \begin{align*}2x+3=27\!\\ {\;} \quad \ \ 2x=24\!\\ {\;} \qquad \ x=12\end{align*}
2. \begin{align*}3x+1=19\!\\ {\;}\quad \ \ 3x=18\!\\ {\;}\quad \ \ \ x=6\end{align*}
2. \begin{align*}6x-13=2x+11 \qquad \qquad 3y+21=90^\circ \!\\ {\;}\ \ \ \ \ 4x=24 \qquad \qquad \qquad \qquad \ 3y=69^\circ \!\\ {\;}\qquad \ 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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