## Math 7 Chapter 2 Lesson 2: Two triangles are congruent

## 1. Summary of theory

### 1.1. Define

Two congruent triangles are two triangles where the three sides of one triangle are equal to the three sides of the other triangle, and the three angles opposite the three sides of this triangle are equal to the three angles opposite the three sides of that triangle.

### 1.2. Sign

To denote the congruence of triangle ABC and triangle A’B’C’ we write:

\(\Delta ABC = \Delta A’B’C’\)

### 1.3. Covenant

When the parallel symbol of two triangles is equal, the letters indicating the names of the corresponding vertices are written in the same order.

\(\Delta ABC = \Delta A’B’C’ \Leftrightarrow \left\{ \begin{array}{l}AB = A’B’,\,AC = A’C’,\,BC = B’ C’\\\widehat A = \widehat {A’},\,\,\widehat B = \widehat {B’},\,\widehat C = \widehat {C’}\end{array} \right. \)

## 2. Illustrated exercise

**Question 1: **Let \(\Delta ABC = \Delta DMN\)

a. Write the above equality in some other form

b. Let AB = 3cm, AC = 4cm, MN=5cm. Calculate the perimeter of each of the above triangles. Any comments?

**Solution guide**

a. Write the equality \(\Delta ABC = \Delta DMN\) in some other form:

\(\begin{array}{l}\Delta ACB = \Delta DNM,\Delta BAC = \Delta DMN,\Delta BCA = \Delta MND\\\Delta CAB = \Delta NDM,\Delta CBA = \Delta NMD \end{array}\)

b. \(\Delta ABC = \Delta DMN \Rightarrow AB = DM,AC = DN,MN = BC.\)

Since AB=3cm, AC=4cm, MN=5cm So:

Perimeter \(\Delta ABC\) is equal to AB+BC+CA=3+4+5=12 (cm)

Perimeter \(\Delta PMN\) is equal to PM+MN+ND=3+4+5=12 (cm)

Comment: Two congruent triangles have the same perimeter.

**Verse 2:** Let \(\Delta ABC = \Delta MNO\). Knowing \(\widehat A = {55^0},\widehat N = {75^0}\) computes the remaining angles of each triangle:

**Solution guide**

\(\Delta ABC = \Delta MNO\) has \(\widehat A = {55^0},\widehat N = {75^0}\)

Thus \(\widehat M = \widehat A = {55^0},\widehat B = \widehat N = {75^0}\)

\(\widehat C = {180^0}(\widehat A + \widehat B) = {180^0} – ({55^0} + {75^0}) = {50^0}\)

So \(\widehat O = \widehat C = {50^0}\)

**Question 3:** Given two congruent triangles ABC and DEG

a. Know \(\widehat A = {20^0},\widehat C = {60^0}\) and \(\widehat E = {100^0}\)

Find the measure of the remaining angles of each triangle.

b. Knowing DG = 5cm can find the length of which side of triangle ABC?

**Solution guide**

a. Since \(\Delta ABC = \Delta DEG\) then:

\(\widehat A = \widehat D = {20^0},\widehat B = {100^0};\,\widehat C = \widehat G = {60^0}\)

b. Easy to see AC=DG=5cm

So the length of side AC=5cm can be found.

## 3. Practice

### 3.1. Essay exercises

**Question 1: **Triangle ABC is equal to triangle PQR, know \(\widehat A = {50^0},\widehat R = {70^0},\widehat B = {60^0},\) side PQ = 6cm, AC = 5cm.

Determine the magnitudes of the remaining angles and the lengths of the sides AB and PR of those two triangles.

**Verse 2: **Let ABC be a triangle with perimeter 21 cm. Side length is three consecutive odd numbers and AB < BC < CA.

Find the lengths of the sides of triangle PQR if triangle ABC is equal to triangle PQR.

**Question 3: **Let ABC be a triangle with M on side BC such that \(\Delta AMB = \Delta AMC.\) Prove that:

a. M is the midpoint of BC

b. AM is the bisector of the origin \(\widehat A\)

c. \(AM \bot BC\)

### 3.2. Multiple choice exercises

**Question 1: **Let \(\Delta ABC = \Delta D{\rm{EF}}\). Know \(\widehat A + \widehat B = {130^0};\widehat E = {55^0}\). calculate angles \(\widehat A;\widehat C;\widehat D;\widehat F\)

A. \(\widehat A = \widehat D = {75^0};\widehat C = \widehat F = {60^0}\)

B. \(\widehat A = \widehat D = {50^0};\widehat C = \widehat F = {75^0}\)

C. \(\widehat A = \widehat D = {75^0};\widehat C = \widehat F = {50^0}\)

D. \(\widehat A = \widehat D = {75^0};\widehat C = \widehat F = {55^0}\)

**Verse 2: **Let \(\Delta ABC = \Delta D{\rm{EF}}\). Know \(\widehat A = {33^0}\). Then

A. \(\widehat D = {33^0}\)

B. \(\widehat D = {42^0}\)

C. \(\widehat E = {33^0}\)

D. \(\widehat D = {66^0}\)

**Question 3: **Given two triangles ABC and DEF with AB = EF, BC = FD, AC = ED, \(\widehat A = \widehat E;\widehat B = \widehat F;\widehat D = \widehat C\) . Then

A. \(\Delta ABC = \Delta D{\rm{EF}}\)

B. \(\Delta ABC = \Delta {\rm{EFD}}\)

C. \(\Delta ABC = \Delta {\rm{EDF}}\)

D. \(\Delta ACB = \Delta {\rm{EFD}}\)

**Question 4: **Let \(\Delta ABC = \Delta D{\rm{EF}}\). Know \(\widehat A = {32^0};\widehat F = {78^0}\). Calculate \(\widehat B;\widehat E\)

A. \(\widehat B = \widehat E = {50^0}\)

B. \(\widehat B = \widehat E = {60^0}\)

C. \(\widehat B = \widehat E = {78^0}\)

D. \(\widehat B = \widehat E = {70^0}\)

**Question 5: **Let \(\Delta ABC = \Delta MNP\). Know AB = 5cm, MP = 7cm and the perimeter of triangle ABC is 22cm. Calculate the remaining sides of each triangle

A. NP = BC = 9cm

B. NP = BC = 11cm

C. NP = BC = 10cm

D. NP = 9cm; BC = 10cm

**Question 6: **Let \(\Delta ABC = \Delta D{\rm{EF}}\). Know that AB = 6cm, AC = 8cm and EF = 10cm. Perimeter of triangle DEF is

A. 24cm

B. 20cm

C. 18cm

D. 30cm

## 4. Conclusion

Through this lesson, two congruent triangles, students should understand the following contents:

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