## Math 8 Chapter 3 Lesson 8: Congruent cases of right triangles

## 1. Theoretical Summary

### 1.1. Similar cases of triangles and right triangles

Two right-angled triangles are similar if:

- This right triangle has an acute angle equal to the acute angle of the other right triangle
- This right triangle has two right angles that are proportional to the two right angles of the other right triangle.

### 1.2. A special sign that identifies two congruent right triangles

If the hypotenuse and one right angled side of one right triangle are proportional to the hypotenuse and one right angle side of another right triangle, then the two right triangles are similar.

### 1.3. Extend

If two triangles are similar then:

- The ratio of two corresponding altitudes is equal to the similarity ratio.
- The ratio of two corresponding bisectors is equal to the ratio of similarity.
- The ratio of two corresponding medians is equal to the similarity ratio.
- The ratio of perimeters is equal to the ratio of similarity.
- The ratio of areas is equal to the square of the similarity ratio.

## 2. Illustrated exercise

### 2.1. Exercise 1

Show pairs of similar triangles in Figure 47.

**Solution guide**

Two right triangles \(ΔDEF\) and \(ΔD’E’F’\) have

\(\dfrac{{DE}}{{DF}} = \dfrac{{D’E’}}{{D’F’}} = \dfrac{1}{2}\)

\(⇒ ΔDEF \) is similar to \(ΔDE’F’\) (two sides of a right angle are proportional)

Applying Py – ta – go theorem to right triangles \(A’B’C’\) and \(ABC\) we get:

\(\eqalign{

& A’C{‘^2} = B’C{‘^2} – A’B{‘^2} \cr

& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {5^2} – {2^2} = 21 \cr

& \Rightarrow A’C’ = \sqrt {21} \cr

& A{C^2} = B{C^2} – A{B^2} \cr

& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {10^2} – {4^2} = 84 \cr

& \Rightarrow AC = \sqrt {84}=2\sqrt{21} \cr} \)

Two right triangles \(ABC\) and \(A’B’C’\) have

\(\dfrac{{AB}}{{A’B’}} = \dfrac{{AC}}{{A’C’}} = 2\)

\( \Rightarrow ΔABC\) congruent \(ΔA’B’C’ \) (two sides are proportional)

### 2.2. Exercise 2

Triangle ABC has side lengths \(3cm, 4cm, 5cm\). Triangle A’B’C’ is similar to triangle ABC and has area \(54c{m^2}\)

Calculate the length of the side length of triangle \(A’B’C’\).

**Solution guide**

Consider \(∆ABC\) with \(AB=3cm,AC=4cm,BC=5cm\).

We have:

\({3^2} + {4^2} = 25 = {5^2} \Rightarrow \Delta ABC\) square at \(A\) (reverse Pythagorean theorem)

So \({S_{ABC}} = \dfrac{1}{2}AB.AC = \dfrac{1}{2}.3.4 = 6c{m^2}\)

Because \(∆ABC ∽ ∆A’B’C’\) (gt)

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

\( \Rightarrow \dfrac{S_{ABC}}{S_{A’B’C’}} = {\left( {\dfrac{{AB}}{{A’B’}}} \right)^2 }\) (area ratio is equal to square of similarity ratio)

Hence: \( \dfrac{6}{54} = {\left( {\dfrac{{AB}}{{A’B’}}} \right)^2}\)

\(\eqalign{

& \Rightarrow {\left( {{{AB} \over {A’B’}}} \right)^2} = {1 \over 9} \cr

& \Rightarrow {{AB} \over {A’B’}} = {1 \over 3} \cr

& \Rightarrow A’B’ = 3AB = 3.3 = 9cm \cr} \)

That is, the length of each side of the triangle \(A’B’C’\) is \(3\) times the length of each side of the side of the triangle \(ABC\).

So the three sides of triangle \(A’B’C’\) are \(A’B’=9cm,A’C’= 12cm, \)\(\,B’C’=15cm\).

## 3. Practice

### 3.1. Essay exercises

**Question 1: **Let \(ABC\) triangle right at \(A,\) \(AC = 9cm, BC = 24cm.\) The perpendicular bisector of \(BC\) intersects the line \(AC\) at \(D\) ), cut \(BC\) at \(M\) (h.30). Calculate the length of the line segment \(CD.\)

**Verse 2: **Let a square trapezoid \(ABCD\) (\(\widehat A = \widehat D = 90^\circ \)) \(AB = 6cm, CD = 12cm,\) \(AD = 17cm.\) On side \ (AD,\) set the line segment \(AE = 8cm\) (h.31). Prove \(\widehat {BEC}= 90^o\).

**Question 3: **Let \(ABC\) triangle right at \(A, \;AC = 4cm, BC = 6cm.\) Draw ray \(Cx\) perpendicular to \(BC\) (ray \(Cx\) and point \(A\) is on the opposite side of the line \(BC\)).Take on the ray \(Cx\) the point \(D\) such that \(BD = 9cm\) (h.32)

Prove that \(BD // AC.\)

**Question 4: **In Figure 33, show similar triangles. Write pairs of similar triangles in the order of their corresponding vertices and explain why they are similar.

### 3.2. Multiple choice exercises

**Question 1:** If two triangles are similar then: Choose the incorrect statement from the following statements?

A. The ratio of two corresponding altitudes is equal to the similarity ratio.

B. The ratio of two corresponding bisectors is equal to the ratio of similarity.

C. The ratio of two corresponding medians is equal to the ratio of similarity.

D. The ratio of perimeters is 2 times the ratio of similarity.

**Verse 2: **Given triangle ABC, right angled at A, foot of altitude AH of triangle ABC divides hypotenuse BC into two segments BH=4cm and HC=9cm. Calculate the area of triangle ABC

A. 39cm^{2}

B. 36cm^{2}

C. 78cm^{2}

D. 18cm^{2}

**Question 3:** Let ABC be a right triangle at A with BC=25 and \(\frac{AB}{AC}=\frac{3}{4}\). Calculating AB, AC

A. AB=16; AC=15

B. AB=15; AC=20

C. AB=10; AC=12

D. AB=20; AC=15

**Question 4: **Given an acute triangle ABC, altitudes BD and CE intersect at H. Prove:

A. \(BH.BD-CH.CE=BC^{2}\)

B. \(CH.CE-BH.BD=BC^{2}\)

C. \(BH.BD-CH.CE=2BC^{2}\)

D. Both a,b,c are wrong

**Question 5: **Let an equilateral triangle DEF be inscribed in an equilateral triangle ABC such that \(DE \perp BC\). The ratio of the areas of triangle DEF and triangle ABC is:

A. \(\frac{1}{6}\)

B. \(\frac{1}{4}\)

C. \(\frac{1}{3}\)

D. \(\frac{2}{5}\)

## 4. Conclusion

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

- Understand the congruent cases of right triangles
- Using the proof that two right-angled triangles are similar, calculate the ratio of altitudes and ratios of areas

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