## Math 8 Chapter 3 Lesson 5: The first similar case

## 1. Theoretical Summary

### 1.1. Theorem

*** Theorem: **If the three sides of one triangle are proportional to the three sides of another triangle, then the two triangles are similar

\(\frac{{AB}}{{A’B’}} = \frac{{BC}}{{B’C’}} = \frac{{CA}}{{C’A’}} \” Rightarrow \Delta ABC \sim \Delta A’B’C’\)

This homologous case is recorded in summary (ccc).

### 1.2. Apply

**Example 1**

Triangle ABC right at A has AB = 6cm, AC = 8cm and triangle A’B’C’ right at A’ has A’B’= 9cm, B’C’ = 15cm. Are two right triangles ABC and A’B’C’ similar?

**Solution guide:**

Applying the Pythagorean theorem, the hypotenuse BC = 10cm and the right angle side A’C’ = 12cm. from there we have:

\(\frac{{AB}}{{A’B’}} = \frac{{BC}}{{B’C’}} = \frac{{CA}}{{C’A’}}\ ,\,\left( { = \frac{2}{3}} \right)\)

So triangle ABC is similar to triangle A’B’C’.

**Example 2**

Are the two triangles whose sides are the following lengths similar?

a) 4cm, 5cm, 6cm and 8mm, 10mm, 12mm

b) 3cm, 4cm, 6cm and 9cm, 15cm, 18cm

**Solution guide:**

a) Two triangles are similar because \(\frac{{40}}{8} = \frac{{50}}{{10}} = \frac{{60}}{{12}}\, \,\left( { = 5} \right)\)

b) The two triangles are not similar because \(\frac{3}{9} \ne \frac{4}{{15}}\)

## 2. Illustrated exercise

### 2.1. Exercise 1

Two triangles \(ABC\) and \(A’B’C’\) have dimensions as shown in figure 32 (with the same unit of measurement as centimeters)

On the sides \(AB\) and \(AC\) of triangle \(ABC\) take two points \(M, N\) respectively, such that

\(AM = A’B’ = 2cm; AN = A’C’ = 3cm\).

Calculate the length of the line segment \(MN\).

Any comments on the relationship between triangles \(ABC, AMN, A’B’C’\)?

**Solution guide**

\(\dfrac{{AM}}{{AB}} = \dfrac{{AN}}{{AC}} = \dfrac{1}{2}\)

\(⇒ MN // BC\) (Inverse Tall theorem)

\(\eqalign{& \Rightarrow {{AM} \over {AB}} = {{AN} \over {AC}} = {{MN} \over {BC}} = {1 \over 2} \cr & \Rightarrow MN = {1 \over 2}BC = {1 \over 2}.8 = 4 \,cm\cr} \)

**Comment:**

\(ΔAMN\) is the same as \(ΔABC\) (because MN//BC); \(ΔAMN = ΔA’B’C'(ccc)\) so \(ΔAMN\) is the same \(ΔA’B’C’\)

Then: \(ΔABC\) is similar to \(ΔA’B’C’\) (same as \(ΔAMN\))

### 2.2. Exercise 2

Let \(ABC\) and \(A’B’C’\) triangles have dimensions as shown in Figure 35.

a) Are triangles \(ABC\) and \(A’B’C’\) similar? Why?

b) Calculate the ratio of the perimeters of the two triangles.

__Solution guide__

a) We have:

\(\begin{array}{l}

\dfrac{{AB}}{{A’B’}} = \dfrac{6}{4} = \dfrac{3}{2};\,\,\dfrac{{AC}}{{A’C ‘}} = \dfrac{9}{6} = \dfrac{3}{2};\\\dfrac{{BC}}{{B’C’}} = \dfrac{{12}}{8} = \dfrac{3}{2}\\

\Rightarrow \dfrac{{AB}}{{A’B’}} = \dfrac{{AC}}{{A’C’}} = \dfrac{{BC}}{{B’C’}} = \dfrac{3}{2}

\end{array}\)

\(\Rightarrow \Delta ABC \text{ same }\Delta A’B’C’\) \(\left( ccc \right)\)

b) Applying the property of the series of equal ratios, we have:

\(\dfrac{{AB}}{{A’B’}} = \dfrac{{AC}}{{A’C’}} = \dfrac{{BC}}{{B’C’}}\’ )\(\, = \dfrac{{AB + AC + BC}}{{A’B’ + A’C’ + B’C’}}\) \( = \dfrac{{{C_{ABC}} }}{{{C_{A’B’C’}}}} = \dfrac{3}{2}\)

(where \(C_{ABC}\) and \(C_{A’B’C’}\) are the perimeters of the two triangles \(ABC, A’B’C’)\)

So the ratio of the perimeter of triangle ABC and the perimeter of triangle A’B’C’ is \(\dfrac{3}{2}\)

### 2.3. Exercise 3

Triangle \(ABC\) has side lengths \(AB = 3cm, AC = 5cm, BC = 7cm\). The triangle \(A’B’C’\) is similar to the triangle \(ABC\) and has a perimeter equal to \(55 cm\).

Calculate the side lengths of \(A’B’C’\) (round to second decimal).

**Solution guide**

\( \Rightarrow \Delta ABC \) is the same \( \Delta A’B’C’\left( {gt} \right)\)

Applying the property of the series of equal ratios, we have:

\(\dfrac{{AB}}{{A’B’}} = \dfrac{{AC}}{{A’C’}} = \dfrac{{BC}}{{B’C’}}\’ )\(\, = \dfrac{{AB + AC + BC}}{{A’B’ + A’C’ + B’C’}}\) \( = \dfrac{{{C_{ABC}} }}{{{C_{A’B’C’}}}}\)

or \(\dfrac{3}{A’B’}\) = \(\dfrac{7}{B’C’}\) = \(\dfrac{5}{A’C’}\) = \ (\dfrac{C_{ABC}}{55}\) = \(\dfrac{3 + 7 + 5}{55}\) = \(\dfrac{{15}}{{55}}\) = \ (\dfrac{3}{11}\)

(where \(C_{ABC}\) and \(C_{A’B’C’}\) are the perimeters of the two triangles \(ABC, A’B’C’\))

\( + )\,\,\dfrac{3}{{A’B’}} = \dfrac{3}{{11}}\)\(\; \Rightarrow A’B’ = \dfrac{{3.11 }}{3} = 11\,cm\)

\( + )\,\,\dfrac{7}{{B’C’}} = \dfrac{3}{{11}}\)\(\; \Rightarrow B’C’ = \dfrac{{7.11 }}{3} \approx 25.67\,cm\)

\( + )\,\,\dfrac{5}{{A’C’}} = \dfrac{3}{{11}}\)\(\; \Rightarrow A’C’ = \dfrac{{5.11 }}{3} \approx 18.33\,cm\)

## 3. Practice

### 3.1. Essay exercises

**Question 1:** Are the two triangles whose sides are the following lengths similar?

a) \(4cm, 5cm, 6cm\) and \(8mm, 10mm, 12mm\);

b) \(3cm, 4cm, 6cm\) and \(9cm, 15cm, 18cm;\)

c) \(1dm, 2dm, 2dm\) and \(1dm, 1dm, 0.5dm\).

**Verse 2:** Right triangle \(ABC\) (\(\widehat A = 90^\circ \)) has \(AB = 6cm, AC = 8cm\) and right triangle \(A’B’C’\) (\ (\widehat {A’} = 90^\circ \)) has \(A’B’ = 9cm, B’C’ = 15cm.\)

Are the two right triangles \(ABC\) and \(A’B’C’\) similar to each other? Why?

**Question 3:** The triangle \(ABC\) has three medians intersecting at \(O.\) Let \(P, Q, R\) be the midpoints of the lines \(OA, OB, OC.\) respectively.

Prove that triangle \(PQR\) is similar to triangle \(ABC.\)

**Question 4: **The triangle \(ABC\) has three acute angles and the orthocenter is the point \(H.\) Let \(K, M, N\) be the midpoints of the lines \(AH, BH, CH, respectively. \)

Prove that triangle \(KMN\) is similar to triangle \(ABC\) with similarity ratio \(\displaystyle k = {1 \over 2}\).

### 3.2. Multiple choice exercises

**Question 1: **Choose the wrong answer. Two triangles ABC and MNP have \(\widehat{BAC}=\widehat{NMP}=90^{\circ}\), AB=3cm, BC=5cm, MN=6cm, MP=8cm, then we can prove:

A. \(\Delta ABC\sim \Delta MNP\)

B. \(\widehat{ACB}=\widehat{MPN}\)

C. Both a and b are wrong

D. Both a and b are correct

**Verse 2: **\(\Delta ABC\sim \Delta EDC\), AB=3; AC=2; CD=3.5; DE=6; BC=x;CE=y. The ratio of the lengths x,y is:

A. 7

B. \(\frac{7}{2}\)

C. \(\frac{7}{4}\)

D. \(\frac{7}{16}\)

**Question 3: **Given triangle ABC, AB=8; BC=7; CA=6.extend side BC a segment CP so that triangle PAB is similar to triangle PCA. The length of side PC is:

A. 7

B. 8

C. 9

D. 10

**Question 4: **Parallel lines L and L’ pass through A and C , perpendicular to BD and divide diagonal BD of rectangle ABCD into three segments of length 1. Area of rectangle rounded to decimal places to be:

A. 4.1

B. 4.2

C. 4.3

D. 4.4

**Question 5:** Let ABC be a triangle. D, E on the sides AB, AC, respectively, such that AD.AC=AE.AB. Prove it!

A. AD.CE=BD.AE

B. \(\frac{BD}{AD}=\frac{CE}{AE}\)

C. DE//BC

D. \(\frac{AD}{BD}=\frac{CE}{AE}\)

## 4. Conclusion

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

- Mastering the content of the theorem about the first similarity case
- Know how to apply the theorem to identify pairs of similar triangles

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