Angle bisectors and the angle bisector theorem

Geometry, grade 10 (4 units) Lesson 17 of 20Learning

Angle bisectors and the angle bisector theorem

The angle bisector as a locus

A locus is all the points that have a certain property – and only them.

Theorem: the angle bisector is the locus of all the points at equal distances from the arms of the angle.

The distance of a point from a line is the length of the perpendicular from it to the line;
the theorem says two things: every point on the angle bisector is at equal distances from the arms (the proof below), and every point at equal distances from the arms is on the angle bisector (Exercise 1).

Given: $\displaystyle AP$ bisects the angle $\displaystyle BAC$ ; $\displaystyle PK\perp AB$ and $\displaystyle PL\perp AC$ .
Prove: $\displaystyle {PK}={PL}$ .

The proof below shows that the triangles $\displaystyle AKP$ and $\displaystyle ALP$ are congruent (ASA).

	Claim	Reason
1	$\displaystyle {\angle KAP}={\angle LAP}$	Given ( $\displaystyle AP$ bisects the angle)
2	$\displaystyle {\angle AKP}={\angle ALP}$	$\displaystyle PK\perp AB$ , $\displaystyle PL\perp AC$ – both angles are right
3	$\displaystyle {\angle APK}={\angle APL}$	(1), (2); the angles of a triangle add up to $\displaystyle 180^{\circ}$
4	$\displaystyle {AP}={AP}$	A common side
5	$\displaystyle \triangle AKP\cong\triangle ALP$	(1), (3), (4); the congruence theorem ASA
6	$\displaystyle {PK}={PL}$	(5); corresponding sides of congruent triangles

Exercise 1

The point $\displaystyle P$ is inside the angle $\displaystyle BAC$ .
$\displaystyle PK\perp AB$ , $\displaystyle PL\perp AC$ , and $\displaystyle {PK}={PL}$ .
Prove: $\displaystyle {\angle KAP}={\angle LAP}$ – that is, $\displaystyle P$ is on the angle bisector.
(A hint: the congruence theorem SSA.)

Where the angle bisectors meet

Theorem: the three angle bisectors of a triangle meet at one point.

$\displaystyle I$ – the point where the bisector of the angle $\displaystyle A$ and the bisector of the angle $\displaystyle B$ cross;
$\displaystyle I$ is on the bisector of $\displaystyle A$ , so at equal distances from $\displaystyle AB$ and from $\displaystyle AC$ ; it is on the bisector of $\displaystyle B$ , so at equal distances from $\displaystyle AB$ and from $\displaystyle BC$ ;
so it is at equal distances from $\displaystyle AC$ and from $\displaystyle BC$ – and it is on the bisector of the angle $\displaystyle C$ .

The point $\displaystyle I$ is at equal distances from the three sides.

The angle bisector theorem

Theorem: an interior angle bisector of a triangle divides the side opposite the angle into two segments whose ratio is 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 $\displaystyle ABC$ .
Prove: $\displaystyle {\frac{BD}{DC}}={\frac{AB}{AC}}$ .

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

	Claim	Reason
1	$\displaystyle {\angle BAD}={\angle DAC}$	Given
2	$\displaystyle CE\parallel AD$	By 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 sides lie opposite equal angles
7	$\displaystyle {\frac{BD}{DC}}={\frac{BA}{AE}}$	(2); a line parallel to the side $\displaystyle CE$ of the triangle $\displaystyle BEC$ – similar triangles
8	$\displaystyle {\frac{BD}{DC}}={\frac{AB}{AC}}$	(6), (7)

Exercise 2

$\displaystyle AD$ 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 theorem

Converse theorem: a line through a vertex of a triangle that divides the side opposite it internally in the ratio of the other two sides is the bisector of the angle at that vertex.

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

Exercise 3

In the triangle $\displaystyle ABC$ , $\displaystyle {AB}={12}$ and $\displaystyle {AC}={8}$ .
The point $\displaystyle D$ is on the side $\displaystyle BC$ , $\displaystyle {BD}={6}$ and $\displaystyle {DC}={4}$ .
Does $\displaystyle AD$ bisect the angle $\displaystyle A$ ?
Explain.

Exercise 4

Prove the angle bisector theorem a second way, by areas.
$\displaystyle AH$ is the altitude to the side $\displaystyle BC$ ; $\displaystyle DG\perp AB$ and $\displaystyle DF\perp AC$ .

Write the areas of $\displaystyle ABD$ and of $\displaystyle ACD$ with $\displaystyle BD$ , $\displaystyle DC$ and the common altitude $\displaystyle AH$ , and find their ratio.
Write the same areas with $\displaystyle AB$ and $\displaystyle DG$ , and $\displaystyle AC$ and $\displaystyle DF$ , and find the ratio again.
What do you know about $\displaystyle DG$ and $\displaystyle DF$ , when $\displaystyle D$ is on the angle bisector?
Compare the two ratios. What have you proved?

Formulas: geometry, grade 10 (4 units)

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Analytic geometry

The distance between the points $\displaystyle (x_1,y_1)$ and $\displaystyle (x_2,y_2)$
$\displaystyle d$ $\displaystyle =$ $\displaystyle \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$
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
$\displaystyle m$ $\displaystyle =$ $\displaystyle \frac{y_2-y_1}{x_2-x_1}$ $\displaystyle x_1\neq x_2$
A line’s equation in explicit form – $\displaystyle m$ the slope, $\displaystyle b$ where it meets the $\displaystyle y$ -axis
$\displaystyle y=mx+b$
A line’s equation in implicit form
$\displaystyle Ax+By+C=0$
The line of slope $\displaystyle m$ through the point $\displaystyle (x_1,y_1)$
$\displaystyle y-y_1$ $\displaystyle =$ $\displaystyle m(x-x_1)$
Parallel lines
$\displaystyle m_1$ $\displaystyle =$ $\displaystyle m_2$
Perpendicular lines
$\displaystyle m_1\cdot m_2$ $\displaystyle =$ $\displaystyle -1$
The slope – $\displaystyle \alpha$ the angle between the line and the positive direction of the $\displaystyle x$ -axis
$\displaystyle m$ $\displaystyle =$ $\displaystyle \tan\alpha$

Similar triangles and midsegments

The similarity ratio – corresponding sides
$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=k$
The ratio of the perimeters
$\displaystyle \frac{P_1}{P_2}$ $\displaystyle =$ $\displaystyle k$
The ratio of the areas
$\displaystyle \frac{S_1}{S_2}$ $\displaystyle =$ $\displaystyle k^{2}$
The ratio of corresponding altitudes, medians and angle bisectors
$\displaystyle \frac{h_1}{h_2}=\frac{m_1}{m_2}=\frac{d_1}{d_2}=k$
A triangle’s midsegment – $\displaystyle DE$ parallel to $\displaystyle BC$
$\displaystyle DE$ $\displaystyle =$ $\displaystyle \frac{1}{2}BC$
A trapezoid’s midsegment – $\displaystyle a$ , $\displaystyle b$ the bases
$\displaystyle EF$ $\displaystyle =$ $\displaystyle \frac{a+b}{2}$

Trigonometry

Sine – $\displaystyle a$ the side opposite the angle, $\displaystyle c$ the hypotenuse
$\displaystyle \sin\alpha$ $\displaystyle =$ $\displaystyle \frac{a}{c}$
Cosine – $\displaystyle b$ the side adjacent to the angle
$\displaystyle \cos\alpha$ $\displaystyle =$ $\displaystyle \frac{b}{c}$
Tangent – the opposite side over the adjacent side
$\displaystyle \tan\alpha$ $\displaystyle =$ $\displaystyle \frac{a}{b}$
The tangent from the sine and the cosine
$\displaystyle \tan\alpha$ $\displaystyle =$ $\displaystyle \frac{\sin\alpha}{\cos\alpha}$
The identity from Pythagoras
$\displaystyle \sin^{2}\alpha+\cos^{2}\alpha$ $\displaystyle =$ $\displaystyle 1$
Angles whose sum is $\displaystyle 90^{\circ}$
$\displaystyle \sin(90^{\circ}-\alpha)$ $\displaystyle =$ $\displaystyle \cos\alpha$
$\displaystyle \cos(90^{\circ}-\alpha)$ $\displaystyle =$ $\displaystyle \sin\alpha$
The sine of an obtuse angle
$\displaystyle \sin(180^{\circ}-\alpha)$ $\displaystyle =$ $\displaystyle \sin\alpha$
A triangle’s area – $\displaystyle \gamma$ the angle between the sides $\displaystyle a$ and $\displaystyle b$
$\displaystyle S$ $\displaystyle =$ $\displaystyle \frac{1}{2}ab\sin\gamma$

Exact values

$\displaystyle \sin30^{\circ}=\cos60^{\circ}=\frac{1}{2}$
$\displaystyle \sin45^{\circ}=\cos45^{\circ}=\frac{\sqrt{2}}{2}$
$\displaystyle \sin60^{\circ}=\cos30^{\circ}=\frac{\sqrt{3}}{2}$
$\displaystyle \tan30^{\circ}$ $\displaystyle =$ $\displaystyle \frac{\sqrt{3}}{3}$
$\displaystyle \tan45^{\circ}$ $\displaystyle =$ $\displaystyle 1$
$\displaystyle \tan60^{\circ}$ $\displaystyle =$ $\displaystyle \sqrt{3}$

Special lines in a triangle

Where the medians meet, $\displaystyle G$ divides the median $\displaystyle AD$ in the ratio $\displaystyle 2:1$
$\displaystyle AG$ $\displaystyle =$ $\displaystyle 2GD$
Where the medians meet, in coordinates
$\displaystyle G\left(\frac{x_1+x_2+x_3}{3},\ \frac{y_1+y_2+y_3}{3}\right)$
The median $\displaystyle AD$ splits the triangle into two triangles of equal area
$\displaystyle S_{ABD}$ $\displaystyle =$ $\displaystyle S_{ACD}$
The angle bisector theorem – $\displaystyle AD$ bisects the angle $\displaystyle A$
$\displaystyle \frac{BD}{DC}$ $\displaystyle =$ $\displaystyle \frac{AB}{AC}$