Similar triangles: the similarity theorems and proofs

Geometry, grade 10 (4 units) Lesson 5 of 20Learning

Similar triangles: the similarity theorems and proofs

Similar triangles

Two triangles are similar when their corresponding angles are equal and their corresponding sides are in the same ratio. That ratio is the similarity ratio, $\displaystyle k$ :

\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=k

$\displaystyle k>1$ – the first triangle is an enlargement of the second; $\displaystyle k<1$ – a reduction; $\displaystyle {k}={1}$ – the triangles are congruent;
we write $\displaystyle \triangle ABC\sim\triangle DEF$ , read “the triangle $\displaystyle ABC$ is similar to the triangle $\displaystyle DEF$ ”.

The similarity theorems

To prove two triangles similar, one of the three similarity theorems is enough:

AA – two angles of one triangle are equal to two angles of the other;
SAS – two sides of one triangle are in the same ratio as two sides of the other, and the angles between them are equal;
SSS – the three sides of one triangle are in the same ratio as the three sides of the other.

The similarity then gives what is not yet known: the other corresponding angles are equal, and the other corresponding sides are in the same ratio.

The order of the letters

The equal angles:

\displaystyle {\angle A}={\angle D}

\displaystyle {\angle B}={\angle E}

\displaystyle {\angle C}={\angle F}

In $\displaystyle \triangle ABC\sim\triangle DEF$ the letters are written by the correspondence: the first goes with the first, the second with the second, the third with the third. So the writing itself tells which sides correspond:

\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}

A segment parallel to a side

Given: in the triangle $\displaystyle ABC$ , the point $\displaystyle E$ is on $\displaystyle AB$ and the point $\displaystyle F$ on $\displaystyle AC$ , and $\displaystyle EF\parallel BC$ .
Prove: $\displaystyle \triangle AEF\sim\triangle ABC$ .

The proof below finds two pairs of equal angles – corresponding angles between the parallel lines – and uses the AA theorem.

	Claim	Reason
1	$\displaystyle EF\parallel BC$	Given
2	$\displaystyle {\angle AEF}={\angle ABC}$	(1); corresponding angles between parallel lines are equal
3	$\displaystyle {\angle AFE}={\angle ACB}$	(1); corresponding angles between parallel lines are equal
4	$\displaystyle \triangle AEF\sim\triangle ABC$	(2), (3); the AA similarity theorem

So $\displaystyle {\frac{AE}{AB}}={\frac{AF}{AC}}={\frac{EF}{BC}}$ : a segment parallel to a side of a triangle cuts off a similar triangle.

Exercise 1

In the triangle $\displaystyle ABC$ , the point $\displaystyle D$ is on $\displaystyle AB$ and the point $\displaystyle E$ on $\displaystyle AC$ .
$\displaystyle {AD}={\frac{1}{3}AB}$ and $\displaystyle {AE}={\frac{1}{3}AC}$ .
Prove: $\displaystyle \triangle ADE\sim\triangle ABC$ .

Exercise 2

The triangle $\displaystyle ABC$ is isosceles: $\displaystyle {AC}={BC}$ .
The point $\displaystyle D$ is on $\displaystyle AC$ and the point $\displaystyle E$ on $\displaystyle BC$ .
$\displaystyle {CD}={\frac{1}{3}AC}$ and $\displaystyle {CE}={\frac{1}{3}BC}$ .
Prove that $\displaystyle ADEB$ is a trapezoid: $\displaystyle DE\parallel AB$ .

Exercise 3

In the triangle $\displaystyle ABC$ , the point $\displaystyle D$ is on $\displaystyle AB$ and the point $\displaystyle E$ on $\displaystyle AC$ , and $\displaystyle DE\parallel BC$ .
$\displaystyle {AD}={4}$ , $\displaystyle {DB}={6}$ and $\displaystyle {DE}={5}$ .
Find $\displaystyle BC$ .

/ – 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

Exercise 4

In the triangle $\displaystyle ABC$ the angle $\displaystyle C$ is right.
$\displaystyle CD$ is the altitude to the hypotenuse $\displaystyle AB$ .
Prove: $\displaystyle \triangle ADC\sim\triangle ACB$ .

Exercise 5

In the right-angled triangle of Exercise 4: $\displaystyle {AD}={2}$ and $\displaystyle {AB}={8}$ .
Find $\displaystyle AC$ .
(A hint: from the similarity, $\displaystyle {\frac{AD}{AC}}={\frac{AC}{AB}}$ .)

/ – 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 6

The sides of one triangle are $\displaystyle 4$ , $\displaystyle 6$ and $\displaystyle 8$ long.
The sides of a second triangle are $\displaystyle 10$ , $\displaystyle 15$ and $\displaystyle 20$ long.
Are the triangles similar?
If they are – by which theorem, and what is the similarity ratio?

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 $\displaystyle A$
$\displaystyle \frac{BD}{DC}$ $\displaystyle =$ $\displaystyle \frac{AB}{AC}$