## Math 8 Chapter 3 Lesson 3: Properties of the bisector of a triangle

## 1. Theoretical Summary

### 1.1. Theorem

The internal bisector of a triangle divides the opposite side into two segments proportional to the two adjacent sides.

The exterior bisector at a vertex of a triangle divides the opposite side into two segments proportional to the two adjacent sides.

\(\begin{array}{l}\frac{{DB}}{{DC}} = \frac{{AB}}{{AC}}\\\frac{{EB}}{{EC}} = \frac{{AB}}{{AC}}\end{array}\)

Thus, the legs of the interior and exterior bisectors of an angle at a vertex of a triangle are the points of the interior and exterior divisions of the opposite side in a ratio equal to the ratio of the respective sides.

\(\frac{{DB}}{{DC}} = \frac{{EB}}{{EC}} = \frac{{AB}}{{AC}}.\)

### 1.2. Attention

The theorem still holds true for the bisector of the exterior angle of the triangle

AE’ is the bisector of angle BAx ( AB AC )

We have: \(\frac{{AB}}{{AC}} = \frac{{E’B}}{{E’C}}\) or \(\frac{{E’B}}{{ AB}} = \frac{{E’C}}{{AC}}\)

## 2. Illustrated exercise

### 2.1. Exercise 1

See figure 23a.

a) Calculate \(\dfrac{x}{y}\)

b) Calculate \(x\) when \(y = 5\)

__Solution guide__

a) Applying the property of the bisector of the triangle to \(\Delta ABC\), the bisector \(AD\) we have:

\(\eqalign{

& {{AB} \over {AC}} = {{BD} \over {DC}} \cr

& \Rightarrow {{3,5} \over {7,5}} = {x \over y} \Rightarrow {x \over y} = {7 \over {15}} \cr} \)

b) When \(y = 5\)

\(\Rightarrow x = 5.\dfrac{7}{{15}} = \dfrac{7}{3}\)

### 2.2. Exercise 2

Calculate \(x\) in figure 24 and round the result to the first decimal place.

a) b)

__Solution guide__

a) \(AD\) is the bisector of angle \(A\) of \(∆ABC\) (gt), so applying the properties of the bisector in the triangle we have:

\(\dfrac{BD}{AB} = \dfrac{DC}{AC}\)

\(\Rightarrow DC = \dfrac{BD.AC}{AB}= \dfrac{3,5.7,2}{4,5}\)

\(\Rightarrow x = 5.6\)

b) \(PQ\) is the angle bisector \(P\) of \(∆PMN\) (gt) so

\(\dfrac{MQ}{MP}= \dfrac{NQ}{NP}\) (property of the bisector of a triangle)

Or \(\dfrac{MQ}{6,2} = \dfrac{x}{8,7}\)

Yes: \(MN=MQ+x=12.5\)

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

\(\Rightarrow \dfrac{x}{8,7} = \dfrac{MQ}{6,2} = \dfrac{x + MQ}{8,7+ 6.2}= \dfrac{12,5} {14.9}\)

\( \Rightarrow x = \dfrac{{12,5.8,7}}{{14.9}} \approx 7.3\)

### 2.3. Exercise 3

The triangle \(ABC\) has side lengths \(AB= m, AC= n\) and \(AD\) is the bisector. Prove that the ratio of the area of triangle \(ABD\) to the area of triangle \(ACD\) is equal to \(\dfrac{m}{n}\).

__Solution guide__

Draw \(AH ⊥ BC\)

We have:

\({S_{ABD}} = \dfrac{1}{2}AH.BD\)

\({S_{ACD}} = \dfrac{1}{2}AH.DC\)

\( \Rightarrow \dfrac{S_{ABD}}{S_{ACD}} = \dfrac{\dfrac{1}{2}AH.BD}{\dfrac{1}{2}AH.DC} = \dfrac {BD}{DC}\)

Otherwise: \(AD\) is the bisector of \(∆ABC\) (gt)

\( \Rightarrow \dfrac{BD}{DC}= \dfrac{AB}{AC} = \dfrac{m}{n}\) (property of the bisector of a triangle)

So \(\dfrac{S_{ABD}}{S_{ACD}} = \dfrac{m}{n}\) (which has to be proven).

## 3. Practice

### 3.1. Essay exercises

**Question 1: **Triangle \(ABC\) has \(AB = 15cm, AC = 20cm,\) \( BC = 25cm.\) The angle bisector \(BAC\) intersects side \(BC\) at \(D\) (h.14)

a) Calculate the length of the line segments \(DB\) and \(DC\).

b) Calculate the ratio of the areas of the two triangles \(ABD\) and \(ACD.\)

**Verse 2: **The triangle \(ABC\) has the bisectors \(AD, BE\) and \(CF\) (h15).

Prove that:

\(\displaystyle {{DB} \over {DC}}. {{EC} \over {EA}}. {{FA} \over {FB}} = 1\)

**Question 3: **An isosceles triangle \(BAC\) has \(BA = BC = a, AC = b.\) The angle bisector \(A\) intersects \(BC\) at \(M\), the angle bisector \ (C\) cuts \(BA\) at \(N\) (h16).

a) Prove that: \(MN // AC.\)

b) Calculate \(MN\) in terms of \(a, b\).

**Question 4: **The triangle \(ABC\) has \(AB = 12cm, AC = 20cm,\) \(BC = 28cm.\) The angle bisector \(A\) intersects \(BC\) at \(D.\) Through \(D\) draw \(DE // AB\) (\(E\) belonging to \(AC\)) (h17).

a) Calculate the length of the line segments \(BD, DC\) and \(DE\).

b) Given that the area of triangle \(ABC\) is \(S\), calculate the area of triangles \(ABD, ADE\) and \(DCE.\)

### 3.2. Multiple choice exercises

**Question 1: **Let ABC be a triangle. The interior bisector of angle A intersects BC at D. Let AB=6, AC=x,BD=9, BC=21. Choose the correct answer of length x:

A. x=14

B. x=12

C. x=8

D. Another result

**Verse 2: **Let ABC be a triangle with AD being the exterior bisector of the angle AD on the line BC. The facts are given in the figure. Which of the following measures is the length of the segment CD?

A. DC=10

B. DC=15

C. DC=6

D. DC=8

**Question 3: **Triangle ABC has side AB=15cm, AC=20cm, BC=25cm. The bisector of angle BAC intersects side BC at D. So the length of DB is:

A. 10

B. \(10\frac{5}{7}\)

C. 14

D. \(14\frac{2}{7}\)

**Question 4: **Triangle ABC has AB=15cm,AC=20cm,BC=25cm. The bisector of angle BAC intersects BC at D. So the ratio of the areas of two triangles ABD and ACD is:

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

B. \(\frac{1}{2}\)

C. \(\frac{3}{4}\)

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

**Question 5: **The lengths of the sides of triangle BAC is proportional to 2:3:4.BD is the internal bisector of the shortest side AC,Dividing AC into two segments AD and CD.If the length of AC is 10, then the length of the segment The longer straight lines of the segments AD and CD are:

A.3.5

B.5

C. \(\frac{40}{7}\)

D.6

## 4. Practice

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

- Know the properties of the bisector of a triangle
- Use the property of bisectors in a triangle to:
- Calculate the length of the line segment
- Set up a series of equal line segments

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