## Math 7 Chapter 1 Lesson 5: Euclidean axiom about parallel lines

## 1. Theory

### 1.1. Euclidean axiom

Through a point outside a line there is only one line parallel to that line.

**Eg:** Figure above: Through a point M outside of line a, there is only one line b parallel to line a.

### 1.2. Properties of two parallel lines

If a line intersects two parallel lines then:

- Two staggered internal angles are congruent.
- Two isotopic angles are congruent.
- Two interior angles on the same side are complementary.

## 2. Illustrated exercise

**Question 1:** Two parallel lines x’x and y’y are intersected by a tangent at two points A and B. Let At be the bisector of \(\widehat {xAB}.\)

a. Does ray At intersect the line y’y? Why?

b. Given \(\widehat {xAB} = {80^0}.\) Calculate \(\widehat {ACB}.\)

**Solution guide**

a. Suppose we At does not cut y’y

Infer AC//y’y. According to Euclidean axiom, AC coincides with x’x. This is absurd so ray At must intersect y’y at C.

b. We have:

\(\widehat {xAt} = \frac{1}{2}\widehat {xAB} = \frac{1}{2}{.80^0} = {40^0}\) (At is the bisector of the ray. of \(\widehat {xAB}\)).

where \(\widehat {xAt} = \widehat {ACB}\) (staggered in)

So \(\widehat {ACB} = {40^0}.\)

**Verse 2: **Given the figure below, know \(\widehat A = {50^0}\) and \(\widehat B = {140^0}\), Ax // By’. Prove that \(\widehat {AOB} = {90^0}.\)

**Solution guide**

Passing O through the line Oz // Ax, we have: \(\widehat {AOz} = \widehat {xAO} = 50{}^0\)(staggered angle in)

Again: \(\widehat {OBy} = {150^0}\)

\( \Rightarrow \widehat {OBy} = {180^0} – {140^0} = {40^0}\)

\(Oz//Ax \Rightarrow Oz//By\)

\( \Rightarrow \widehat {BOz’} = \widehat {OBy} = {40^0}\) (staggered angle in)

Thus: \(\widehat {AOz} = \widehat {z’OB} = {50^0} + {40^0} = {90^0}\) or \(\widehat {AOB} = {90^ 0}.\)

**Question 3: **Given the figure below, know Ax // By. Prove that \(\widehat A + \widehat B + \widehat C = {360^0}.\)

**Solution guide**

Draw through C the line Cz // Ax we have:

\(\widehat A + \widehat {ACz} = {180^0}\) (complementary innermost angles)

Again: \(Cz//Ax \Rightarrow Cz//By \Rightarrow \widehat B + \widehat {zCB} = {180^0}\) (complementary innermost angles)

\( \Rightarrow \widehat A + \widehat B + \widehat {ACz} + \widehat {zCB} = {360^0}\,\,\,\,or\,\,\,\,\,\widehat A + \widehat B + \widehat C = {360^0}.\)

## 3. Practice

### 3.1. Essay exercises

**Question 1: **Given an angle xOy whose measure is \({30^0}\). A point A belongs to Ox. Through A construct ray A’y // Oy and lie in angle xOy.

a. Calculate OAy’.

b. Let Ot and At’ be the bisectors of the angles xOy and xAy, respectively. Show that Ot//At’.

**Verse 2:** Let \(xOy = {120^0}\) and \(Ot\) be the bisector of that angle. On the ray Oy take the point A, through A draw the line At’ // Ot.

a. Calculate angle yAt’

b. From A draw a line Ax’ parallel to Ox. Compare two angles t’Ax’ and tOx.

### 3.2. Multiple choice exercises

**Question 1: **Find the wrong answer. If a // b then:

A. \(\widehat {{A_2}} = \widehat {{B_2}}\)

B. \(\widehat {{A_2}} = \widehat {{B_1}}\)

C. \(\widehat {{A_1}} + \widehat {{B_3}} = {180^o}\)

D. \(\widehat {{A_4}} + \widehat {{B_4}} = {180^o}\)

**Verse 2: **Given the figure, know AB // CD

Calculate the measure \(\widehat A,\widehat B\)

A. \(\widehat A = {105^o},\,\,\widehat B = {130^o}\)

B. \(\widehat A = {115^o},\,\,\widehat B = {130^o}\)

C. \(\widehat A = {115^o},\,\,\widehat B = {150^o}\)

D. \(\widehat A = {135^o},\,\,\widehat B = {115^o}\)

**Question 3: **Given the figure, know a // b.

The measure of angle AOB is:

A. 50^{o}

B. 70^{o}

C. 90^{o}

D. 120^{o}

**Question 4: **Given the following figure, know x // y and \(\widehat {{M_1}} = {55^o}\). Calculate measure \(\widehat {{N_1}}\)

A. 55^{o}

B. 35^{o}

C. 60^{o}

D. 125^{o}

## 4. Conclusion

Through this lesson, you should know the following:

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