## Math 7 Chapter 3 Lesson 1: The relationship between the angle and the opposite side in a triangle

## 1. Summary of theory

**Theorem 1: **In a triangle, the angle opposite the larger side is the larger angle.

**Example 1:** Triangle ABC if AC > AB then \(\widehat B > \widehat C\)

**Theorem 2: **In a triangle, the side opposite the larger angle is the larger side.

**Example 2: **ABC is a triangle so \(\widehat B > \widehat C\) then AC > AB

**Comment:**

- In \(\Delta ABC\,,\,\,AC > AB \Leftrightarrow \widehat B > \widehat C\)
- In an obtuse triangle (or right triangle), the obtuse angle (or right angle) is the largest angle, so the side opposite the obtuse angle (or right angle) is the largest side.

## 2. Illustrated exercise

**Question 1:** Let \(\Delta ABC\) square at A (AB < AC).

The bisector of \(\widehat B\) intersects AC at D, through C draw a perpendicular to AC intersecting the opposite ray of DB at I. Prove AB < CI; AC < CI.

**Solution guide**

We have: \(\widehat {{B_1}} = \widehat {{B_2}}\) (BD is the bisector \(\widehat {ABC}\)) where AB // CI (the same perpendicular to AC)

\( \Rightarrow \widehat {{B_1}} = \widehat {{I_1}}\) (staggered in)

\( \Rightarrow \widehat {{B_2}} = \widehat {I\,}.\)

So \(\Delta BCI\) is balanced at A with AB < BC.

So AB < CI.

Similarly, we can also prove:

AC < CI (because \(AC < BC \Rightarrow AC < CI\))

**Verse 2: **Let ABC be a triangle. On the opposite ray of ray BA we take some point D different from point B and on the opposite ray of ray CA we take a point E such that CE = BD. Prove that BC is less than DE.

**Solution guide**

Considering \(\Delta ACD.\) Angle DCE is the exterior angle of the vertex C of that triangle, so we have:

\(\widehat {DCE} > \widehat {CDA}\)

Two triangles BCD and EDC have two equal sides each pair.

BD = EC (assumption)

CD is the common edge

The two angles included between the two sides are not equal \(\widehat {DCE} > \widehat {CDB}\) so the two sides opposite the two angles are not equal.

We infer: BC < DE.

**Question 3: **Let \(\Delta ABC\) have AC > AB, M is the mid point of BC. Connect AM. On the opposite ray of MA take D such that MA = MD. Connect BD. Compare \(\widehat {BAM}\) and \(\widehat {CAM}\).

**Solution guide**

\(\Delta AMC\) and \(\Delta DMB\) have:

AM = DM (gt)

MC = BM (gt)

\(\widehat {AMC} = \widehat {DMB}\) (opposite)

So \(\Delta AMC = \Delta DMB\,\,(cgc)\)

So \(\widehat {CAM} = \widehat {BDM}\) and AC = DB.

Where AC > AB (gt)

\( \Rightarrow DB > AB\)

In \(\Delta ABD\) there is \(BD > AB \Rightarrow \widehat {BAM} > \widehat {BDM}\)

Or \(\widehat {BAM} > \widehat {CAM}\)

## 3. Practice

### 3.1. Essay exercises

**Question 1:** Let \(\Delta ABC\) be balanced at A. Take point E on segment BC, take point F on segment BC extending, point D on AC extending towards C. Join AE, AF, BD. Prove:

a. AB < AE

b. AB < AF

c. BD > BC

**Verse 2: **Let \(\Delta ABC\) have 3 sides satisfying the relation AC > CB > BA. Let I be the intersection of the internal bisectors of \(\widehat B\) and \(\widehat A\). Prove that IB < IA < IC.

**Question 3:** Given \(\Delta ABC\)(AC > AB), M is the mid point of BC. On AB and AC take 2 points P and N such that BP = CN. Prove: \(\widehat {APN} = \widehat {ANP}\)

### 3.2. Multiple choice exercises

**Question 1: **Let ABC be a triangle with \(\widehat C > \widehat B\) (angle B and angle C are acute angles). Draw the bisector AD. Compare BD and CD

A. Not qualified to compare

B. BD = CD

C. BD < CD

D. BD > CD

**Verse 2: **Let ABC be a triangle with an obtuse angle A. Take point E on side AB, take point F on side AC. Choose the correct answer.

A. BF > EF

B. EF < BC

C. BF < BC

D. Both A, B, C are correct

**Question 3: **Let ABC be a triangle with AB < AC. Let M be the mid point of BC. On the opposite carriage of the MA take point D such that MA = MD. Compare CDA angle and CAD angle?

A. \(\widehat {CA{\rm{D}}} > \widehat {CDA}\)

B. \(\widehat {CA{\rm{D}}} = \widehat {CDA}\)

C. \(\widehat {CA{\rm{D}}} < \widehat {CDA}\)

D. \(\widehat {CA{\rm{D}}} = 2. \widehat {CDA}\)

**Question 4: **Let ABC be a triangle with AB + AC = 10cm, AC – AB = 4cm. Compare angle B and angle C

A. \(\widehat C < \widehat B\)

B. \(\widehat C > \widehat B\)

C. \(\widehat C = \widehat B\)

D. \(\widehat C = 2. \widehat B\)

**Question 5: **Let ABC be a triangle with \(\widehat A = {80^0},\widehat B – \widehat C = {20^0}\). Please choose the best answer:

A. AC < AB < BC

B. AB < AC < BC

C. BC < AC < AB

D. AC < BC < AB

## 4. Conclusion

Through this lesson, you should know the following:

- Understand the relationship between the angle and the opposite side in a triangle.
- Apply to solve related problems.

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