The angle bisector theorem

Geometry, grade 10 (5 units) Lesson 15 of 26Learning

The angle bisector theorem

The angle bisector theorem: An internal angle bisector of a triangle divides the opposite side into two segments whose ratio equals the ratio of the sides enclosing the angle.

\displaystyle \frac{BD}{DC}

\displaystyle =

\displaystyle \frac{AB}{AC}

Given: $\displaystyle AD$ bisects the angle $\displaystyle A$ of the triangle $\displaystyle ABC$ , and $\displaystyle D$ is on $\displaystyle BC$ .
To prove: $\displaystyle {\frac{BD}{DC}}={\frac{AB}{AC}}$ .

The proof below draws through $\displaystyle C$ a line parallel to $\displaystyle AD$ , to the point $\displaystyle E$ on the extension of the side $\displaystyle BA$ . It shows that $\displaystyle {AE}={AC}$ , and then uses Thales’ theorem in the triangle $\displaystyle BEC$ .

	Claim	Reason
1	$\displaystyle {\angle BAD}={\angle DAC}$	Given
2	$\displaystyle CE\parallel AD$	Construction
3	$\displaystyle {\angle ACE}={\angle DAC}$	(2); alternate angles between parallel lines
4	$\displaystyle {\angle AEC}={\angle BAD}$	(2); corresponding angles between parallel lines
5	$\displaystyle {\angle ACE}={\angle AEC}$	(1), (3), (4)
6	$\displaystyle {AE}={AC}$	(5); in a triangle, equal angles are opposite equal sides
7	$\displaystyle {\frac{BD}{DC}}={\frac{BA}{AE}}$	(2); Thales’ theorem in the triangle $\displaystyle BEC$
8	$\displaystyle {\frac{BD}{DC}}={\frac{AB}{AC}}$	(6), (7)

Exercise 1

$\displaystyle AD$ bisects the angle $\displaystyle A$ of the triangle $\displaystyle ABC$ , and $\displaystyle D$ is on the side $\displaystyle BC$ .

$\displaystyle {AB}={6}$ , $\displaystyle {AC}={9}$ and $\displaystyle {BC}={10}$ .
Find $\displaystyle BD$ .
(A hint: $\displaystyle {BD}={x}$ , and then $\displaystyle {DC}={10-x}$ .)

$\displaystyle {AB}={8}$ , $\displaystyle {BD}={4}$ and $\displaystyle {DC}={6}$ .
Find $\displaystyle AC$ .

The converse of the angle bisector theorem

The converse: A line through a vertex of a triangle that divides the opposite side internally in the ratio of the other two sides bisects the angle at that vertex.

In the triangle $\displaystyle ABC$ : if $\displaystyle D$ is on $\displaystyle BC$ and $\displaystyle {\frac{BD}{DC}}={\frac{AB}{AC}}$ , then $\displaystyle AD$ bisects the angle $\displaystyle A$ .

The proof below shows that the bisector of the angle $\displaystyle A$ reaches $\displaystyle BC$ exactly at $\displaystyle D$ :

the bisector of the angle $\displaystyle A$ meets $\displaystyle BC$ at a point $\displaystyle D'$ ;
by the angle bisector theorem $\displaystyle {\frac{BD'}{D'C}}={\frac{AB}{AC}}$ , so $\displaystyle {\frac{BD'}{D'C}}={\frac{BD}{DC}}$ ;
adding $\displaystyle 1$ to both sides: $\displaystyle {\frac{BC}{D'C}}={\frac{BC}{DC}}$ , so $\displaystyle {D'C}={DC}$ ;
$\displaystyle D'$ and $\displaystyle D$ are both on the segment $\displaystyle BC$ , at the same distance from $\displaystyle C$ – they are one point, and $\displaystyle AD$ is the bisector.

Exercise 2

In the triangle $\displaystyle ABC$ , $\displaystyle D$ is on the side $\displaystyle BC$ .
$\displaystyle {\frac{BD}{DC}}={\frac{AB}{AC}}$ .
Prove: $\displaystyle {\angle BAD}={\angle DAC}$ .

Exercise 3

$\displaystyle AD$ bisects the angle $\displaystyle A$ of the triangle $\displaystyle ABC$ , and $\displaystyle D$ is on the side $\displaystyle BC$ .
$\displaystyle E$ is on the side $\displaystyle AC$ , and $\displaystyle DE\parallel AB$ .
Prove: $\displaystyle {AE}={DE}$ .

Exercise 4

In the triangle $\displaystyle ABC$ , $\displaystyle {AB}={12}$ , $\displaystyle {AC}={8}$ and $\displaystyle {BC}={10}$ .
$\displaystyle AD$ bisects the angle $\displaystyle A$ , and $\displaystyle D$ is on $\displaystyle BC$ .
$\displaystyle BE$ bisects the angle $\displaystyle B$ , and $\displaystyle E$ is on $\displaystyle AC$ .
The bisectors cross at $\displaystyle I$ .

Find

\displaystyle BD

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Find the ratio

\displaystyle \frac{AI}{ID}

. (A hint:

\displaystyle BI

bisects an angle of the triangle

\displaystyle ABD

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 5

In the triangle $\displaystyle ABC$ , $\displaystyle {AB}={AC}$ , and $\displaystyle AD$ bisects the angle $\displaystyle A$ .
What does the angle bisector theorem say about $\displaystyle D$ ?
Which theorem about the isosceles triangle does this agree with?

Formulas: geometry, grade 10 (5 units)

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Introduction to analytic geometry

The distance between the points $\displaystyle (x_1,y_1)$ and $\displaystyle (x_2,y_2)$
$\displaystyle d$ $\displaystyle =$ $\displaystyle \sqrt{(x_1-x_2)^{2}+(y_1-y_2)^{2}}$
The midpoint of the segment
$\displaystyle M\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)$
The slope of a line through two points
$\displaystyle m$ $\displaystyle =$ $\displaystyle \frac{y_2-y_1}{x_2-x_1}$ $\displaystyle x_1\neq x_2$
The line of slope $\displaystyle m$ through the point $\displaystyle (x_1,y_1)$
$\displaystyle y-y_1$ $\displaystyle =$ $\displaystyle m(x-x_1)$
The explicit equation of a line – $\displaystyle m$ the slope, $\displaystyle b$ where it crosses the $\displaystyle y$ -axis
$\displaystyle y=mx+b$
The general equation of a line
$\displaystyle Ax+By+C=0$
Parallel lines
$\displaystyle m_1$ $\displaystyle =$ $\displaystyle m_2$
Perpendicular lines
$\displaystyle m_1\cdot m_2$ $\displaystyle =$ $\displaystyle -1$
A canonical circle – its centre at the origin, $\displaystyle R$ its radius
$\displaystyle x^{2}+y^{2}=R^{2}$
The circle with centre $\displaystyle (a,b)$ and radius $\displaystyle R$
$\displaystyle (x-a)^{2}+(y-b)^{2}=R^{2}$

Midsegments and the meeting point of the medians

A triangle’s midsegment – $\displaystyle DE$ is parallel to $\displaystyle BC$
$\displaystyle DE$ $\displaystyle =$ $\displaystyle \frac{1}{2}BC$
A trapezoid’s midsegment – $\displaystyle a$ , $\displaystyle b$ the bases
$\displaystyle EF$ $\displaystyle =$ $\displaystyle \frac{a+b}{2}$
The meeting point of the medians $\displaystyle G$ divides the median $\displaystyle AD$ in the ratio $\displaystyle 2:1$
$\displaystyle AG$ $\displaystyle =$ $\displaystyle 2GD$
Pythagoras’ theorem and its converse: the triangle is right-angled, the right angle opposite $\displaystyle c$
$\displaystyle a^{2}+b^{2}=c^{2}$

Proportion and similarity

A property of a proportion
$\displaystyle \frac{a}{b}=\frac{c}{d}\ \Rightarrow\ \frac{a+b}{b}=\frac{c+d}{d}$
Thales’ theorem – $\displaystyle DE\parallel BC$
$\displaystyle \frac{AD}{DB}$ $\displaystyle =$ $\displaystyle \frac{AE}{EC}$
The extended Thales theorem – $\displaystyle DE\parallel BC$
$\displaystyle \frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC}$
The angle bisector theorem – $\displaystyle AD$ bisects the angle $\displaystyle A$
$\displaystyle \frac{BD}{DC}$ $\displaystyle =$ $\displaystyle \frac{AB}{AC}$
The similarity ratio – corresponding sides
$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=k$
Heights, medians, perimeters, the radii of the circumscribed and of the inscribed circles
$\displaystyle \frac{h_1}{h_2}=\frac{m_1}{m_2}=\frac{P_1}{P_2}=\frac{R_1}{R_2}=\frac{r_1}{r_2}=k$
The ratio of the areas
$\displaystyle \frac{S_1}{S_2}$ $\displaystyle =$ $\displaystyle k^{2}$

The circle

An inscribed angle is half the central angle on the same arc
$\displaystyle \angle ACB$ $\displaystyle =$ $\displaystyle \frac{1}{2}\angle AOB$
An inscribed angle on the diameter $\displaystyle AB$
$\displaystyle \angle ACB=90^{\circ}$
A quadrilateral $\displaystyle ABCD$ inscribed in a circle – opposite angles
$\displaystyle \angle A+\angle C$ $\displaystyle =$ $\displaystyle 180^{\circ}$
The angle between the tangent $\displaystyle AT$ and the chord $\displaystyle AB$ equals an inscribed angle on $\displaystyle AB$ from its other side
$\displaystyle \angle BAT$ $\displaystyle =$ $\displaystyle \angle ACB$
Two tangents from the point $\displaystyle P$
$\displaystyle PA=PB$
A quadrilateral $\displaystyle ABCD$ circumscribing a circle
$\displaystyle AB+CD$ $\displaystyle =$ $\displaystyle BC+AD$
The chords $\displaystyle AB$ and $\displaystyle CD$ meet at the point $\displaystyle E$
$\displaystyle AE\cdot EB$ $\displaystyle =$ $\displaystyle CE\cdot ED$
Two secants from the point $\displaystyle P$
$\displaystyle PA\cdot PB$ $\displaystyle =$ $\displaystyle PC\cdot PD$
A tangent $\displaystyle PT$ and a secant $\displaystyle PAB$ from the point $\displaystyle P$
$\displaystyle PT^{2}$ $\displaystyle =$ $\displaystyle PA\cdot PB$