## Math 6 Chapter 2 Lesson 6: Bisector of angle

## 1. Summary of theory Tóm

The bisector of an angle is the ray that lies between the two sides of the angle and forms two equal angles with that side.

– Ray Oz is the bisector of angle xOt, then:

- Oz ray lies between two rays Ox, Oy.
- \(\widehat {xOy} = \widehat {zOy}\)

Or \(\widehat {xOz} = \widehat {zOy} = \dfrac{1}{2}\widehat {xOy}.\)

**Attention:**

Each angle has only one bisector.

The line containing the bisector of an angle is the bisector of that angle.

## 2. Illustrated exercise

**Question 1:** Draw the bisector of the right angle.

**Solution guide**

- \(\widehat {xOy}\) is a flat angle, so \(\widehat {xOy}=180^0\)
- The ray \(Oz\) is the bisector of \(\widehat {xOy}\) so \(\widehat {xOz} = \widehat {zOy} = \dfrac{1}{2}\widehat {xOy}=90 ^0.\)
- The opposite ray of the ray \(Oz\) is also a bisector of the flat angle \(xOy\).

We have the following figure:

**Verse 2: **Let \(\widehat {AOB} = {88^0}\). Ray OC is the bisector of \(\widehat {AOB}\). Calculate the measure \(\widehat {BOC}\)?

**Solution guide**

Since OC is the bisector of \(\widehat {AOB}\)

So \(\widehat {AOB} = \widehat {BOC} = \dfrac{1}{2}\widehat {AOB}.\)

Hence: \(\widehat {BOC} = \dfrac{1}{2}. 88^0=44^0.\)

## 3. Practice

### 3.1. Essay exercises

**Question 1:** Given two adjacent angles \(\widehat {xOy}\) and \(\widehat {yOz}.\) Draw the bisectors Om and On of the angles xOy, yOz. Know \(\widehat {mOn} = {37^0}\). Calculate angle \(\widehat {xOz}?\)

**Verse 2:** Let angle \(\widehat {xOy} = {a^0}\). A ray of Oz lies in the angle \(\widehat {xOy}\). Show that the angle formed by the bisectors of the angles \(\widehat {xOz}\) and \(\widehat {zOy}\) does not depend on the position of the ray Oz in the angle xOy.

**Question 3: **Let three rays OA, OB, OC share the origin O and take in that order. Know \(\widehat {AOB} = {72^0};\,\,\widehat {BOC} = {108^0}\)

a) Prove that the three points A, O, C are collinear.

b) Draw the bisector of angle AOB and in angle BOC, draw ray OE such that \(\widehat {EOD} = {90^0}.\) Calculate angle \(\widehat {BOE}.\)

c) Prove that OE is the bisector of angle \(\widehat {BOC}.\)

### 3.2. Multiple choice exercises Bài

**Question 1: **Let \(\widehat {AOB} = {110^0}\) and \(\widehat {AOC} = {55^0}\) such that \(\widehat {AOB}\) and \(\widehat {AOC }\) are not adjacent. Choose the wrong sentence

A. Ray OC lies between two rays OA and OB

B. Ray OC is the bisector of angle AOB

C. \(\widehat {BOC} = {65^0}\)

D. \(\widehat {BOC} = {55^0}\)

**Verse 2: **Let \({\widehat {xOy}}\) and \({\widehat {yOz}}\) be complementary adjacent angles. Given that \({\widehat {xOy} = {{120}^0}}\) and ray Ot is the bisector of \({\widehat {yOz}}\). Calculate the measure of angle xOt

A. 140^{0}

B. 150^{0}

C. 90^{0}

D. 120^{0}

**Question 3: **Given angle AOB and bisector OC of that angle. Draw the bisector OM of angle BOC. Know \(\widehat {BOM} = {35^0}\). Calculate the measure of angle AOB

A. 150^{0}

B. 120^{0}

C. 140^{0}

D. 160^{0}

**Question 4: **Let the ray On be the bisector of \(\widehat {mOt}\). Given that \(\widehat {mOn} = {70^0}\), the measure of \(\widehat {mOt}\) is

A. 140^{0}

B. 120^{0}

C. 35^{0}

D. 60^{0}

**Question 5: **Let \(\widehat {AOB} = {90^0}\) and ray OB be the bisector of angle AOC. Then angle AOC is

A. Right angle

B. Sharp angle

C. Obtuse angle

D. Flat angle

## 4. Conclusion

Through this lesson, you should be able to understand the following:

- Understand and state the definition of the bisector of an angle.
- Show that a ray is the bisector of an angle in the simple case.
- Calculate the measure of an angle based on the bisector of an angle.
- Draw the bisector of a given angle.

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