## Math 8 Chapter 3 Lesson 6: Second similar case

## 1. Summary of theory

### 1.1. Define

If two sides of one triangle are proportional to two sides of another triangle and the angles formed by those pairs of sides are congruent, then the two triangles are similar.

General: Δ ABC,Δ A’B’C’ has A’B’/AB = A’C’/AC = B’C’/BC ⇒ Δ ABC ∼ Δ A’B’C’

### 1.2. Apply

**Example 1**

Given two triangles \(ABC\) and \(DEF\) with dimensions as shown in Figure 36.

– Compare the ratios \(\dfrac{{AB}}{{DE}}\) and \( \dfrac{{AC}}{{DF}}\)

– Measure the line segments \(BC, EF\). Calculate the ratio \(\dfrac{{BC}}{{EF}}\), compare with the above ratios and predict the similarity of the two triangles \(ABC\) and \(DEF.\)

**Solution guide**

\(\dfrac{{AB}}{{DE}} = \dfrac{{AC}}{{DF}} = \dfrac{1}{2}\)

Measuring the edges we have: \(BC \approx \,3,6 ; EF \approx 7.2 \)

\(\eqalign{& \Rightarrow {{BC} \over {EF}} = {1 \over 2} \cr & \Rightarrow {{AB} \over {DE}} = {{AC} \over {DF} } = {{BC} \over {EF}} = {1 \over 2} \cr} \)

Prediction: \(ΔABC\) is similar to \(ΔDEF\).

**Example 2**

Show pairs of similar triangles from the following triangles (h.38):

**Solution guide**

Consider \(ΔABC\) and \(ΔDEF\) to have:

\(\eqalign{& \widehat A = \widehat D = {70^o} \cr & {{AB} \over {AC}} = {{DE} \over {DF}} = {2 \over 3} \cr} \)

\(⇒ ΔABC\) is the same as \(ΔDEF\) (cgc)

**Example 3**

a) Draw a triangle \(ABC\) with \(\widehat {BAC} = {50^o}\), \(AB = 5cm, AC = 7.5cm\) (h.39)

b) Take on the sides \(AB, AC\) two points \(D, E\) respectively such that \(AD = 3cm, AE = 2cm\). Are the two triangles \(AED\) and \(ABC\) similar to each other? Why ?

**Solution guide**

a) Students draw pictures by themselves

b) We have: \(\dfrac{{AB}}{{AC}} = \dfrac{5}{{7,5}} = \dfrac{2}{3}\)

\(\dfrac{{AE}}{{AD}} = \dfrac{2}{3}\)

Consider \(\Delta ABC\) and \(\Delta AED\) with

\(\widehat A\) common

\(\dfrac{{AB}}{{AC}} = \dfrac{{AE}}{{AD}} = \dfrac{2}{3}\)

\( \Rightarrow \Delta ABC\) same \( \Delta AED\,\,\left( {cgc} \right) \)

## 2. Illustrated exercise

### 2.1. Exercise 1

On one side of angle \(xOy\) (\(\widehat {xOy} \ne {180^0}\)), Place segments \(OA= 5cm, OB= 16cm\). On the second edge of that corner, place segments \(OC= 8cm, OD= 10cm\).

a) Prove that the two triangles \(OCB\) and \(OAD\) are similar.

b) Call the intersection of the sides \(AD\) and \(BC\) \(I\), prove that the two triangles \(IAB\) and \(ICD\) have equal angles each pair one.

**Solution guide**

a) We have:

\(\dfrac{OA}{OC} = \dfrac{5}{8}\) ; \(\dfrac{OD}{OB} = \dfrac{10}{16} = \dfrac{5}{8}\)

\(\Rightarrow \dfrac{OA}{OC} = \dfrac{OD}{OB}\)

Consider \(∆OCB\) and \(∆OAD\) to have:

+) \(\widehat O\) common

+) \(\dfrac{OA}{OC} = \dfrac{OD}{OB}\) (proven above)

\(\Rightarrow OCB \) is the same as \(∆OAD\) ( cgc)

\( \Rightarrow \widehat {ODA} = \widehat {CBO}\) (2 corresponding angles) or \(\widehat{CDI}\) = \(\widehat{IBA}\)

b) Consider \(∆ICD\) and \(∆IAB\) with

\(\widehat{CID}\) = \(\widehat{AIB}\) (two opposite angles) (1)

\(\widehat{CDI}\) = \(\widehat{IBA}\) (according to sentence a) (2)

By the sum of the three angles in a triangle, we have:

\(\eqalign{

& \widehat {CID} + \widehat {CDI} + \widehat {ICD} = {180^0} \cr

& \widehat {AIB}+\widehat {IBA} + \widehat {IAB} = {180^0} \cr} \)

\( \Rightarrow \widehat {CID} + \widehat {CDI} + \widehat {ICD} \) \(= \widehat {AIB}+\widehat {IBA} + \widehat {IAB}\) (3)

From (1), (2) and (3) deduce: \( \widehat {ICD}=\widehat {IAB}\)

So two triangles \(IAB\) and \(ICD\) have equal angles one by one.

### 2.2. Exercise 2

Prove that if triangle \(A’B’C’\) is similar to triangle \(ABC\) by the ratio \(k\), then the ratio of the two medians corresponds to those two triangles. is also equal to \(k\).

__Solution guide__

Assume that \(∆A’B’C’\) is similar to \(∆ABC\) in the ratio \(k, A’M’, AM\) as the two corresponding medians.

Since \(∆A’B’C’\) is similar to \(∆ABC\) (assumption)

\(\dfrac{A’B’}{AB} = \dfrac{B’C’}{BC}\) (property of two similar triangles)

Where \(B’C’ = 2B’M’, BC = 2BM\) (median property)

\( \Rightarrow \dfrac{{A’B’}}{{AB}} = \dfrac{{2B’M’}}{{2BM}} = \dfrac{{B’M’}}{{BM} }\)

Consider \(∆ABM\) and \( ∆A’B’M’\) with:

\(\widehat{B} = \widehat{B’}\) (because \(∆A’B’C’\) is similar to \(∆ABC\))

\( \dfrac{{A’B’}}{{AB}} = \dfrac{{B’M’}}{{BM}}\) (proven above)

\( \Rightarrow ∆A’B’M’ \) is the same as \(∆ABM\) (cgc)

\( \Rightarrow \dfrac{A’M’}{AM}= \dfrac{A’B’}{AB} = k.\)

## 3. Practice

### 3.1. Essay exercises

**Question 1: **Let \(ABC\) triangle with \(AB = 12cm, AC = 15cm,\) \(BC = 18cm.\)

On the edge \(AB\), place the line segment \(AM = 10cm,\) on the side \(AC\) put the line segment \(AN = 8cm.\) Calculate the length of the line segment \(MN.\)

**Verse 2: **Trapezoid \(ABCD \;(AB // CD)\) has \(AB = 4cm, CD = 16cm\) and \(BD = 8cm \) (h23).

Prove that \(\widehat {BAD} = \widehat {DBC}\) and \(BC = 2 AD.\)

**Question 3: **Let \(ABC\) triangle with \(\widehat A = 60^\circ \), \(AB = 6cm, AC = 9cm\)

a) Construct a triangle similar to triangle \(ABC\) in the ratio of similarity \(\displaystyle k = {1 \over 3}\)

b) Name some other construction methods and draw the figure in each specific case.

**Question 4:** Construct a triangle \(ABC\), knowing \(\widehat{A}={60^o}\) and, the ratio \(\dfrac{AB}{AC} = \dfrac{4}{5}\) and the height \(AH = 6cm\).

### 3.2. Multiple choice exercises

**Question 1:** If two triangles MNP and QRS have \(\frac{MN}{QS}=\frac{MP}{RS}\) and \(\widehat{M}=\widehat{S}\) then:

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

B. \(\Delta MNP \sim \Delta SQR\)

C. \(\Delta MNP \sim \Delta RSQ\)

D. \(\Delta MNP \sim \Delta RSQ\)

**Verse 2: **If two triangles EFH and GKL have \(\frac{EF}{GK}=\frac{EH}{GL}\) and \(\widehat{E}=\widehat{G}\) then:

A. \(\widehat{EHF}=\widehat{GKL}\)

B. \(\widehat{ÈFH}=\widehat{GLK}\)

C. \(\widehat{FEH}=\widehat{LKG}\)

D. \(\widehat{EHF}=\widehat{KLG}\)

**Question 3: **Given triangle ABC, AAD is the bisector. Prove:

A. \(AD^{2}=AB.AC+DB.DC\)

B. \(AD^{2}=BD.DC-AB.AC\)

C. \(AD^{2}=AB.AC-DB.DC\)

D. Both a,b,c are wrong

**Question 4: **Given two triangles ABC, DEF has AB=4cm, AC=6cm, BC=5cm, DE=2cm,EF=1cm, \(\widehat{ABC}=\widehat{DEF}\).Prove:

A. \(\widehat{BAC}=2\widehat{EDF}\)

B. \(\widehat{DFE}=\frac{\widehat{BAC}}{2}+\widehat{ACB}\)

C. Both a and b are correct

D. Both a and b are wrong

**Question 5: **Let \(\Delta MNP \sim \Delta EFH\) in the ratio k, MM’, EE’ be the medians of triangles MNP and EFH, respectively. We can prove it.

A. \(\frac{EE’}{MM’}=k\)

B. \(\frac{MM’}{EE’}=k^{2}\)

C. \(\frac{MM’}{EE’}=k\)

D. \(\frac{EE’}{MM’}=k^{2}\)

## 4. Conclusion

Through this lesson, you will learn some key topics as follows:

- Firmly grasp the content of the theorem (GT, KL), understand how the proof consists of 2 main steps.
- Apply the theorem to identify pairs of similar triangles, do exercises for calculating the lengths of sides and proof exercises.

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