## Math 8 Chapter 3 Lesson 2: The inverse theorem and its consequences of Talet’s theorem

## 1. Theoretical Summary

### 1.1. Island theorem

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

### 1.2. Consequences of Talet’s theorem

If a line intersects the remaining two sides of a triangle and is parallel to the remaining sides, then it forms a new triangle with three corresponding sides proportional to the remaining three sides of the given triangle.

## 2. Illustrated exercise

### 2.1. Exercise 1

The triangle \(ABC\) has \(AB=6cm\); \(AC=9cm\).

Take on the edge \(AB\) the point \(B’\), on the edge \(AC\) the point \(C’\) such that \(AB’=2cm\); \(AC’=3cm\) (h8)

1) Compare the ratios \(\dfrac{{AB’}}{{AB}}\) and \(\dfrac{{AC’}}{{AC}}\).

2) Draw a line \(a\) through \(B’\) and parallel to \(BC\), the line \(a\) intersects \(AC\) at \(C”\) .

a) Calculate the length of the line segment \(AC”\).

b) Any comments about \(C’\) and \(C”\) and about the two lines \(BC\) and \(B’C’\)?

**Solution guide**

first)

\(\begin{array}{l}\dfrac{{AB’}}{{AB}} = \dfrac{2}{6} = \dfrac{1}{3}\\\dfrac{{AC’} }{{AC}} = \dfrac{3}{9} = \dfrac{1}{3}\\ \Rightarrow \dfrac{{AB’}}{{AB}} = \dfrac{{AC’}} {{AC}}\end{array}\)

2)

a) Since \(B’C”//BC\) , according to Talent’s theorem we have:

\(\dfrac{{AB’}}{{AB}} = \dfrac{{AC”}}{{AC}} = \dfrac{1}{3}\)

\( \Rightarrow AC” = \dfrac{1}{3}AC = \dfrac{1}{3}.9 = 3\,cm\)

b) We have: \(AC’ = AC” = 3\,cm \Rightarrow C’ \equiv C”\)

Since \(C’ \equiv C” \Rightarrow B’C’ \equiv B’C”\) \(B’C’//BC\)

### 2.2. Exercise 2

Look at Figure 9.

a) How many pairs of parallel lines are there in the given figure?

b) What is the quadrilateral \(BDEF\)?

c) Compare the ratios \(\dfrac{{AD}}{{AB}};\dfrac{{AE}}{{AC}};\dfrac{{DE}}{{BC}}\) and for commenting on the relationship between the pairs of corresponding sides of the two triangles \(ADE\) and \(ABC\).

**Solution guide**

a) We have:

\(\begin{array}{l}\dfrac{{AD}}{{AB}} = \dfrac{3}{{3 + 6}} = \dfrac{3}{9} = \dfrac{1} {3}\\\dfrac{{AE}}{{AC}} = \dfrac{5}{{5 + 10}} = \dfrac{5}{{15}} = \dfrac{1}{3} \\ \Rightarrow \dfrac{{AD}}{{AB}} = \dfrac{{AE}}{{AC}}\end{array}\)

According to the Inverse Tales theorem, then \(DE//BC\)

\(\begin{array}{l}\dfrac{{CE}}{{CA}} = \dfrac{{10}}{{10 + 5}} = \dfrac{{10}}{{15}} = \dfrac{2}{3}\\\dfrac{{CF}}{{CB}} = \dfrac{{14}}{{14 + 7}} = \dfrac{{14}}{{21} } = \dfrac{2}{3}\\ \Rightarrow \dfrac{{CE}}{{CA}} = \dfrac{{CF}}{{CB}}\end{array}\)

According to the inverse Tales theorem, \(EF//AB\)

In the given figure there are 2 pairs of parallel lines.

b) The quadrilateral \(BDEF\) has \(BD//EF;DE//BF\) so \(BDEF\) is a parallelogram.

c) Since \(BDEF\) is a parallelogram, \(DE = BF = 7\) (Property of a parallelogram).

We have: \(\dfrac{{DE}}{{BC}} = \dfrac{7}{{7 + 14}} = \dfrac{1}{3}\)

Thus: \(\dfrac{{AD}}{{AB}} = \dfrac{{AE}}{{AC}} = \dfrac{{DE}}{{BC}} = \dfrac{1}{ 3}\)

**Comment: **Two triangles \(ADE\) and \(ABC\) have corresponding pairs of proportional sides.

## 3. Practice

### 3.1. Essay exercises

**Question 1: **Given a triangle \(ABC\) with side \(BC = a.\) On the side \(AB\) take the points \(D\) and \(E\) such that \(AD = DE = EB.\ ) From \(D, E\) draw lines parallel to \(BC\), intersecting \(AC\) in order at \(M, N\) (h.5)

Calculate in terms of \(a\) the lengths of the lines \(DM\) and \(EN.\)

**Verse 2: **Figure 7 shows the triangle \(ABC\) right at \(A,\) \(MN // BC, AB = 24cm,\) \(AM = 16cm,\) \(AN = 12cm.\) Calculate degrees length \(x, y\) of the lines \(NC\) and \(BC\).

**Question 3: **Trapezoid \(ABCD\; (AB // CD)\) has two diagonals \(AC\) and \(BD\) intersecting at \(O\) (h.8).

Prove that: \(OA.OD = OB.OC.\)

**Question 4: **Given a trapezoid \(ABCD \;(AB // CD)\). The line parallel to the base \(AB\) intersects the sides and diagonals \(AD, BD, AC\) and \(BC\) at the points \(M, N, P, Q\, respectively) ) (h.9)

Prove that \(MN = PQ.\)

### 3.2. Multiple choice exercises

Question 1: Let ABC be a triangle with centroids G.A’,B’,C’ the midpoints of BC,CA ,AB.Which of the following statements is false.

A. GA and GN are proportional to GA’ and GB’

B. GA and GA’ are proportional to AC and B’C

C. GB’ and GC’ are proportional to GB and GA

D. The altitude AH of triangle ABC and the altitude GD of triangle GBC are proportional to AA’ and GA’.

**Verse 2:** Let M and N be two points on sides AB and AC respectively of triangle ABC. Know MN=6cm,AM=3cm,MB=5cm,AC=16cm,Cn=10cm. The length of the side is:

A. BC=10cm

B. BC=9cm

C. BC=16cm

D. Another result

**Question 3: **Given triangle ABC, M and N are two points on sides AB, AC respectively. Know AM=2cm,MB=5cm,AN=3.2cm,NC=8cm,BC=14cm. Length of line segment MN is:

A. MN=5.6cm

B. MN=4cm

C. MN=8cm

D. MN=5,2cm

**Question 4: **Calculate x in the figure below:

A. x=4.5

B. x=3

C. x=2

D. All 3 sentences above are wrong

**Question 5: **Given the figures shown in the figure, AF and CD are calculated and the answer is:

A. 7.5 and 1.2

B. 3 and 1.85

C. 4.5 and 2

D. All 3 sentences above are wrong

## 4. Conclusion

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

- Grasp the content of Talet’s converse theorem.
- Apply the theorem to determine pairs of parallel lines in the drawing with the given data
- Understand how to prove the consequence of Talett’s theorem, especially understand the possible cases when the line B’C’ is parallel to the side BC
- Through each picture, students can write a formula or a series of equal ratios.1

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