## Math 8 Chapter 3 Lesson 1: Talet’s theorem in a triangle

## 1. Theoretical Summary

### 1.1. Ratio of two line segments

**a) Definition**

The ratio of two line segments is the ratio of their lengths in the same unit of measure.

– The ratio of two lines AB and CD is denoted by \(\dfrac{{AB}}{{C{\text{D}}}}\)

**b) Attention: **The ratio of two line segments does not depend on the unit of measure chosen.

### 1.2. Proportional line segment

Definition: Two lines AB and CD are said to be proportional to two lines A’B’ and C’D’ if they have a

\(\dfrac{{AB}}{{C{\text{D}}}} = \dfrac{{A’B’}}{{C’D’}}\;\;or\;\;\ dfrac{{AB}}{{A’B’}} = \dfrac{{C{\text{D}}}}{{C’D’}}\)

### 1.3. Talet’s theorem in triangles

If a line is parallel to one side of a triangle and intersects the other two sides, then it defines on those two sides proportional line segments.

General : Δ ABC, B’C’//BC; B’ ∈ AB, C’ AC

We have: \(\frac{{AB’}}{{AB}} = \frac{{AC’}}{{AC}};\frac{{AB’}}{{BB’}} = \frac {{AC’}}{{C’C}};\frac{{BB’}}{{AB}} = \frac{{CC’}}{{AC}}\)

## 2. Illustrated exercise

### 2.1. Exercise 1

Draw a triangle \(ABC\) on student paper as shown in Figure 3. Construct a line \(a\) parallel to the side \(BC\), cut the two sides \(AB, AC\) in the order at \(B’\) and \(C’\).

The line \(a\) defines on the edge \(AB\) three lines \(AB’,B’B\) and \(AB\), and defines on the edge \(AC\) three corresponding lines. are \(AC’,C’C\) and \(AC\) respectively.

Compare ratios:

\(a)\,\dfrac{{AB’}}{{AB}}\) and \(\dfrac{{AC’}}{{AC}}\)

\(b)\,\,\dfrac{{AB’}}{{B’B}}\) and \(\dfrac{{AC’}}{{C’C}}\)

\(c)\,\,\dfrac{{B’B}}{{AB}}\) and \(\dfrac{{C’C}}{{AC}}\)

**Solution guide**

\(a)\,\dfrac{{AB’}}{{AB}}=\dfrac{{AC’}}{{AC}}=\dfrac{5}{8}\)

\(b)\,\,\dfrac{{AB’}}{{B’B}}=\dfrac{{AC’}}{{C’C}}=\dfrac{5}{3}\)

\(c)\,\,\dfrac{{B’B}}{{AB}}=\dfrac{{C’C}}{{AC}}=\dfrac{3}{8}\)

### 2.2. Exercise 2

Write the ratio of pairs of line segments of the following lengths:

a) \(AB = 5cm\) and \(CD = 15 cm\)

b) \(EF = 48 cm\) and \(GH = 16 dm\)

c) \(PQ = 1.2m\) and \(MN = 24 cm\)

**Solution guide**

a) We have \(AB = 5cm\) and \(CD = 15 cm\)

\(\Rightarrow \dfrac{AB}{CD}= \dfrac{5}{15}= \dfrac{1}{3}\).

b) \(EF= 48 cm, GH = 16 dm = 160 cm\)

\( \Rightarrow \dfrac{EF}{GH}= \dfrac{48}{160}= \dfrac{3}{10}\)

c) \(PQ= 1.2m = 120cm, MN= 24cm\)

\(\Rightarrow \dfrac{PQ}{MN} = \dfrac{120}{24} = 5.\)

### 2.3. Exercise 3

Given that the length of \(AB\) is \(5\) times the length of \(CD\) and the length of \(A’B’\) is \(12\) times the length of \(CD) \). Calculate the ratio of two lines \(AB\) and \(A’B’\).

**Solution guide**

The length \(AB\) is \(5\) times the length \(CD\) so \(AB= 5CD\).

The length \(A’B’\) is \(12\) times the length \(CD\) so \(A’B’= 12CD\).

\( \Rightarrow \) The ratio of the two lines \(AB\) and \(A’B’\) is:

\(\dfrac{AB}{A’B’}= \dfrac{5CD}{12CD} = \dfrac{5}{12}\)

## 3. Practice

### 3.1. Essay exercises

**Question 1: **Write the ratio of the following pairs of line segments:

a) \(AB = 125cm, CD = 625 cm\);

b) \(EF = 45cm, E’F’ = 13.5dm\);

c) \(MN = 555cm, M’N’ = 999cm\);

d) \(PQ = 10101cm, P’Q’ = 303.03m\).

**Verse 2: **The line segment \(AB\) is five times the line segment \(CD\); the line segment \(A’B’\) is seven times the line segment \(CD\).

a. Calculate the ratio of two lines \(AB\) and \(A’B’\).

b. Given the line segment \(MN = 505cm\) and the line segment \(M’N’ = 707cm\), ask the two line segments \(AB , A’B’\) proportional to the two lines \(MN\) ) and \(M’N’\) or not?

**Question 3: **Given a trapezoid \(ABCD\) with \(AB // CD\) and \(AB < CD\).

The line parallel to the base \(AB\) intersects the sides \(AD, BC\) respectively at \(M\) and \(N.\)

Prove that:

a. \(\displaystyle{{MA} \over {AD}} = {{NB} \over {BC}}\)

b. \(\displaystyle{{MA} \over {MD}} = {{NB} \over {NC}}\)

c. \(\displaystyle{{MD} \over {DA}} = {{NC} \over {CB}}\)

**HD: **Extend the rays \(DA, CB\) that intersect at \(E\) (h.3), apply Taylor’s theorem in the triangle and the property of the proportions to prove.

**Question 4:** Given triangle \(ABC\). From the point \(D\) on the edge \(BC\), draw lines parallel to the sides \(AB\) and \(AC\), they intersect the edges \(AC\) and \(AB\) ) in order at \(F\) and \(E\)

Prove that: \(\displaystyle {{AE} \over {AB}} + {{AF} \over {AC}} = 1\)

### 3.2. Multiple choice exercises

**Question 1:** Based on the factors on the figure, applying Talet’s theorem, we get:

A. x=1

B. x=5

C. x=2

D. x=1.5

**Verse 2: **Indicate a false ratio if we apply Talet’s theorem

A. \(\frac{LC}{CB}=\frac{LK}{LA}\)

B. \(\frac{IB}{IK}=\frac{IA}{ID}\)

C. \(\frac{IB}{ID}=\frac{IA}{IK}\)

D. \(\frac{KA}{KL}=\frac{KD}{KC}\)

**Question 3: **Let \(\frac{AB}{A’B’}=\frac{CD}{C’D’}\) (1)

(1) <=> AB.C’D’=A’B’.CD (I)

(2) <=> \(\frac{AB}{CD}=\frac{A’B’}{C’D’}\) (II)

A.(I) and (II) are both wrong

B.(I) and (II) are both correct

C.Only (I) is correct

D.Only (II) is correct

**Question 4:** For line segments: AB=6cm,CD=4cm,PQ=8cm,EF=10cm,MN=25mm,RS=15mm.Choose the correct statement from the following:

A. Two lines AB and PQ are proportional to two lines EF and RS

B. Two lines AB and RS are proportional to two lines EF and MN

C. Two lines CD and AB are proportional to two lines PQ and EF

D. All 3 sentences above are wrong

**Question 5: **For line segments: AB=8cm,CD=6cm,MN=12mm,PQ=x. Find x so that Ab and CD are proportional to MN and PQ

A. x=18mm

B. x=9cm

C. x=0.9cm

D. Both a,b,c are wrong

Through this lesson, you will learn some of the main topics as follows:

- The ratio of two line segments is the ratio of their lengths in the same unit of measure.
- The ratio of two line segments does not depend on how the unit of measure is chosen (as long as only the same unit of measure is chosen when measuring).
- Students master about the proportional line segment and the content of Talet’s theorem, apply the theorem to find equal ratios on the figure in the textbook.

.

=============

## Leave a Reply