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Perpendicular Bisectors

Intersect line segments at their midpoints and form 90 degree angles with them.

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Perpendicular Bisectors

Imagine an archeologist in Cairo, Egypt, found three bones 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?

Perpendicular Bisectors

Recall that a perpendicular bisector intersects a line segment at its midpoint and is perpendicular. 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.

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

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 PC¯¯¯¯¯¯¯¯,QC¯¯¯¯¯¯¯¯\begin{align*}\overline{PC}, \overline{QC}\end{align*}, and RC¯¯¯¯¯¯¯¯\begin{align*}\overline{RC}\end{align*} are perpendicular bisectors, then LC=MC=OC\begin{align*}LC=MC=OC\end{align*}.

Determining Unknown Values

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

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*}.

Applying the Properties of Perpendicular Bisectors

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

a) Which segments are equal?

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 the perpendicular bisector of MP¯¯¯¯¯¯¯¯¯\begin{align*}\overline{MP}\end{align*}.

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

4x+34xx=11=8=2\begin{align*}4x+3 &= 11 \\ 4x &= 8\\ x &= 2\end{align*}

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

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).

Further Exploration

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?

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

Archaeology 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.

Examples

Example 1

Find the value of x\begin{align*}x\end{align*}. m\begin{align*}m\end{align*} is the perpendicular bisector of AB\begin{align*}AB\end{align*}.

By the Perpendicular Bisector Theorem, both segments are equal. Set up and solve an equation.

x+6x=22=16\begin{align*}x+6 &=22\\ x &=16\end{align*}

Example 2

Determine if ST\begin{align*}\overleftrightarrow{S T}\end{align*} is the perpendicular bisector of XY¯¯¯¯¯¯¯¯\begin{align*}\overline{XY}\end{align*}. Explain why or why not.

2. ST\begin{align*}\overleftrightarrow{S T}\end{align*} is not necessarily the perpendicular bisector of XY¯¯¯¯¯¯¯¯\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 ST\begin{align*}\overleftrightarrow{S T}\end{align*} will extend to make a right angle with XY¯¯¯¯¯¯¯¯\begin{align*}\overline{XY}\end{align*}.

Review

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

For Question 2, determine if ST\begin{align*}\overleftrightarrow{ST}\end{align*} is the perpendicular bisector of XY¯¯¯¯¯¯¯¯\begin{align*}\overline{XY}\end{align*}. Explain why or why not.

For Questions 3-7, consider line segment AB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB}\end{align*} with endpoints A(2,1)\begin{align*}A(2, 1)\end{align*} and B(6,3)\begin{align*}B(6, 3)\end{align*}.

1. Find the slope of AB\begin{align*}AB\end{align*}.
2. Find the midpoint of AB\begin{align*}AB\end{align*}.
3. Find the equation of the perpendicular bisector of AB\begin{align*}AB\end{align*}.
4. Find AB\begin{align*}AB\end{align*}. Simplify the radical, if needed.
5. Plot A,B\begin{align*}A, B\end{align*}, and the perpendicular bisector. Label it m\begin{align*}m\end{align*}. How could you find a point C\begin{align*}C\end{align*} on m\begin{align*}m\end{align*}, such that C\begin{align*}C\end{align*} would be the third vertex of equilateral triangle ABC\begin{align*}\triangle ABC\end{align*}? You do not have to find the coordinates, just describe how you would do it.

For Questions 8-12, consider ABC\begin{align*}\triangle ABC\end{align*} with vertices A(7,6),B(7,2)\begin{align*}A(7, 6), B(7, -2)\end{align*} and C(0,5)\begin{align*}C(0, 5)\end{align*}. Plot this triangle on graph paper.

1. Find the midpoint and slope of AB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB}\end{align*} and use them to draw the perpendicular bisector of AB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB}\end{align*}. You do not need to write the equation.
2. Find the midpoint and slope of BC¯¯¯¯¯¯¯¯\begin{align*}\overline{BC}\end{align*} and use them to draw the perpendicular bisector of BC¯¯¯¯¯¯¯¯\begin{align*}\overline{BC}\end{align*}. You do not need to write the equation.
3. Find the midpoint and slope of AC¯¯¯¯¯¯¯¯\begin{align*}\overline{AC}\end{align*} and use them to draw the perpendicular bisector of AC¯¯¯¯¯¯¯¯\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 11 as the center of your circle.
6. Fill in the blanks: There is exactly _________ circle which contains any __________ points.
7. Fill in the blanks of the proof of the Perpendicular Bisector Theorem.

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

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

Statement Reason
1.
2. D\begin{align*}D\end{align*} is the midpoint of AB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB}\end{align*}
3. Definition of a midpoint
4. CDA\begin{align*}\angle CDA\end{align*} and CDB\begin{align*}\angle CDB\end{align*} are right angles
5. CDACDB\begin{align*}\angle CDA \cong \angle CDB\end{align*}
6. Reflexive PoC
7. CDACDB\begin{align*}\triangle CDA \cong \triangle CDB\end{align*}
8. AC¯¯¯¯¯¯¯¯CB¯¯¯¯¯¯¯¯\begin{align*}\overline{AC} \cong \overline{CB}\end{align*}
1. Write a two column proof.

Given: ABC\begin{align*}\triangle ABC\end{align*} is a right isosceles triangle and BD¯¯¯¯¯¯¯¯\begin{align*}\overline{BD}\end{align*} is the \begin{align*}\bot\end{align*} bisector of AC¯¯¯¯¯¯¯¯\begin{align*}\overline{AC}\end{align*}

Prove: ABD\begin{align*}\triangle ABD\end{align*} and CBD\begin{align*}\triangle CBD\end{align*} are congruent.

To view the Review answers, open this PDF file and look for section 5.2.

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Vocabulary Language: English

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 $90^\circ$.

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.