## Math 7 Chapter 3 Lesson 7: Properties of the perpendicular bisector of a line segment

## 1. Summary of theory

### 1.1. Properties of points on the perpendicular

**Theorem 1: (Forward Theorem): **A point on the perpendicular bisector of a line segment is equidistant from the two endpoints of that segment.

**Eg:** Since point M lies on the perpendicular bisector of AB, MA = MB.

**Theorem 2: (Inverse theorem): **If a point is equidistant from two endpoints of a line segment, then it is on the perpendicular bisector of that segment.

**Eg:** Since MA = MB, point M lies on the perpendicular bisector of line AB.

**Comment: **From the forward and reverse theorems we have: the set of points equidistant from the two endpoints of a line segment is the perpendicular bisector of that line.

### 1.2. Application

Draw the perpendicular bisector of line segment MN using a straightedge and compass; as follows:

- Using M as the center draw an arc with radius greater than \(\frac{1}{2}MN\). Using N as the center draw an arc with the same radius.
- These two arcs have two points in common, P and Q.
- Use a ruler to draw the line PQ. That is the perpendicular bisector of the line segment MN.

## 2. Illustrated exercise

**Question 1:** Triangle ABC has AC > AB, bisector AD. On AC take point E such that AE=AB. Prove that AD is perpendicular to BE.

**Solution guide**

Connect BE and ED

Consider \(\Delta ADB\) and \(\Delta ADE\) with:

AB = AE (gt)

\(\widehat {BAD} = \widehat {EAD}\) (AD is the bisector \(\widehat {BAC}\)).

AD common edge

So \(\Delta ADB = \Delta ADE\,\,(cgc)\)

Derive DB = DE

Again, AB = AE

Hence AB is the perpendicular bisector of BE.

So \(AD \bot BE.\)

**Verse 2: **Given \(\Delta ABC.\) Find a point equidistant from sides AB, AC and equidistant from two vertices A and B.

**Solution guide**

Every point on the bisector of angle A is equidistant from sides AB and AC.

Every point on the perpendicular bisector of AB is equidistant from two points A and B.

So the point M to find is the intersection of the bisector and the above median.

**Question 3: **Let m be the perpendicular bisector of the line segment AB, and C the point on m. Let Cx be the opposite ray of ray CA and Cn be the bisector of angle BCx. Prove that Cn is perpendicular to m.

**Solution guide**

Let H be the intersection of m and AB.

Consider \(\Delta AHC\) and \(\Delta BHC\) with HA = HB (H is the point on the perpendicular bisector of AB)

\(\widehat {AHC} = \widehat {AHC} = {90^0}\)

CH is the common edge

So \(\Delta AHC = \Delta BHC\) (cgc)

So \(\widehat {ACH} = \widehat {BCH}\)

So CH is the bisector of \(\widehat {ACB}\)

Cn is the bisector of \(\widehat {BCx}\) (gt)

So m and Cn are two bisectors of two complementary angles ACB and BCx, so \({C_n} \bot m.\)

## 3. Practice

### 3.1. Essay exercises

**Question 1: **Prove that there are no equidistant, three collinear points.

**Verse 2: **Let the line segment AB belong to the half-plane shore d. Determine the point M in d such that point M is equidistant from two points A and B.

**Question 3: **On the perpendicular bisector d of the line segment AB take the point M. Lower \(MH \bot AB.\) On the segment MH take the point P. Let E be the intersection of AP with MB. Let F be the intersection of BP with MA.

a) Prove that MH is the bisector of angle AMB

b) Prove that MH is the perpendicular bisector of the line segment EF

c) Prove AF = BE.

### 3.2. Multiple choice exercises

**Question 1: **Given a right triangle ABC at A, with \(\widehat C = {30^0}\), the perpendicular bisector of BC intersects AC at M. Choose the correct answer

A. BM is the median of triangle ABC

B. BM = AB

C. BM is the bisector of angle ABC

D. BM is the perpendicular bisector of triangle ABC

**Verse 2: **Given triangle ABC is isosceles at, with \(\widehat A = {40^0}\), the perpendicular bisector AB intersects BC at D. Calculate the measure of angle CAD

A. 30^{0}

B. 45^{0}

C. 60^{0}

D. 40^{0}

**Question 3: **Let O be the intersection of the three perpendicular bisectors in triangle ABC. Then O is:

A. A point equidistant from the three sides of triangle ABC

B. The point is equidistant from the three vertices of the triangle ABC

C. Center of the circumcircle of triangle ABC

D. Answers B and C are correct

**Question 4: **If a triangle has a median and a perpendicular bisector, what kind of triangle is it?

A. Right triangle

B. Isosceles triangle

C. Equilateral triangle

D. Isosceles right triangle

**Question 5: **Let ABC be an isosceles triangle at A. The perpendicular bisector of AC intersects AB at D. Let CD be the bisector of angle ACB. Calculate the angles of triangle ABC

A. \(\widehat A = {30^0},\widehat B = \widehat C = {75^0}\)

B. \(\widehat A = {40^0},\widehat B = \widehat C = {70^0}\)

C. \(\widehat A = {36^0},\widehat B = \widehat C = {72^0}\)

D. \(\widehat A = {70^0},\widehat B = \widehat C = {55^0}\)

## 4. Conclusion

Through this lesson, you should know the following:

- Understand the definition of the perpendicular bisector of a line segment.
- Know the properties of points on the perpendicular bisector.
- Apply related problems.

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