Ratios in similar triangles

Geometry, grade 10 (5 units) Lesson 17 of 26Learning

Ratios in similar triangles

The ratio of corresponding heights

Theorem: In similar triangles, the ratio of corresponding heights equals the similarity ratio.

Given: $\displaystyle \triangle ABC\sim\triangle DEF$ ; $\displaystyle AH$ and $\displaystyle DK$ are the heights to $\displaystyle BC$ and to $\displaystyle EF$ .
To prove: $\displaystyle {\frac{AH}{DK}}={\frac{AB}{DE}}$ .

The proof below finds two smaller similar triangles: $\displaystyle ABH$ and $\displaystyle DEK$ – each has a right angle, and $\displaystyle {\angle B}={\angle E}$ .

	Claim	Reason
1	$\displaystyle \triangle ABC\sim\triangle DEF$	Given
2	$\displaystyle {\angle ABH}={\angle DEK}$	(1); corresponding angles of similar triangles
3	$\displaystyle {\angle AHB}={\angle DKE}={90^{\circ}}$	$\displaystyle AH$ and $\displaystyle DK$ are heights
4	$\displaystyle \triangle ABH\sim\triangle DEK$	(2), (3); AA
5	$\displaystyle {\frac{AH}{DK}}={\frac{AB}{DE}}$	(4); corresponding sides of similar triangles

Medians and angle bisectors

The same holds for the other special lines: in similar triangles, the ratio of corresponding medians and the ratio of corresponding angle bisectors equal the similarity ratio.

Medians: $\displaystyle AM$ and $\displaystyle DN$ cut off the triangles $\displaystyle ABM$ and $\displaystyle DEN$ ; $\displaystyle {\frac{BM}{EN}}={\frac{BC}{EF}}={k}$ , so these are similar by SAS (Exercise 1).
Angle bisectors: they cut off triangles with equal angles – half of $\displaystyle \angle A$ and half of $\displaystyle \angle D$ , and $\displaystyle {\angle B}={\angle E}$ – similar by AA.

Exercise 1

$\displaystyle \triangle ABC\sim\triangle DEF$ .
$\displaystyle M$ is the midpoint of $\displaystyle BC$ , and $\displaystyle N$ the midpoint of $\displaystyle EF$ .
Prove: $\displaystyle {\frac{AM}{DN}}={\frac{AB}{DE}}$ .

The ratio of the perimeters

Theorem: The ratio of the perimeters of similar triangles equals the similarity ratio.

Each side of the first triangle is $\displaystyle k$ times the corresponding side of the second: $\displaystyle {a_1}={ka_2}$ , $\displaystyle {b_1}={kb_2}$ , $\displaystyle {c_1}={kc_2}$ . So

P_1=ka_2+kb_2+kc_2=k(a_2+b_2+c_2)=kP_2

\displaystyle \frac{P_1}{P_2}

\displaystyle =

\displaystyle k

The ratio of the areas

Theorem: The ratio of the areas of similar triangles equals the square of the similarity ratio.

Both the base and the height grow by $\displaystyle k$ : $\displaystyle {BC}={k\cdot EF}$ , and by the theorem on heights $\displaystyle {AH}={k\cdot DK}$ . So

\frac{S_{ABC}}{S_{DEF}}=\frac{\frac{1}{2}BC\cdot AH}{\frac{1}{2}EF\cdot DK}=k\cdot k=k^{2}

\displaystyle \frac{S_1}{S_2}

\displaystyle =

\displaystyle k^{2}

Example: the sides of a triangle are doubled. Its perimeter is doubled too, and its area is $\displaystyle {2^{2}}={4}$ times larger.

Similar polygons

Two polygons are similar when their angles are equal in pairs and their corresponding sides are in one ratio $\displaystyle k$ .

Diagonals from corresponding vertices split them into similar triangles, in the same ratio $\displaystyle k$ ;
so the ratio of their perimeters is $\displaystyle k$ , and the ratio of their areas is $\displaystyle k^{2}$ .

A model and the real thing are similar: a plan of a flat, a photo and its enlargement, a model plane and a plane.

Exercise 2

$\displaystyle \triangle ABC\sim\triangle DEF$ .
$\displaystyle {AB}={6}$ and $\displaystyle {DE}={9}$ .
The perimeter of $\displaystyle ABC$ is $\displaystyle 20$ , and the area of $\displaystyle DEF$ is $\displaystyle 45$ .

Find the perimeter of

\displaystyle DEF

/ – 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 area of

\displaystyle ABC

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Exercise 3

In the triangle $\displaystyle ABC$ , $\displaystyle D$ is on the side $\displaystyle AB$ and $\displaystyle E$ is on the side $\displaystyle AC$ .
$\displaystyle DE\parallel BC$ and $\displaystyle {\frac{AD}{DB}}={\frac{2}{3}}$ .
The area of the triangle $\displaystyle ADE$ is $\displaystyle 8$ .

Find the area of the triangle

\displaystyle ABC

. (A hint: what is the similarity ratio of

\displaystyle ADE

\displaystyle ABC

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Find the area of the trapezoid

\displaystyle DBCE

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Exercise 4

The ratio of the areas of two similar triangles is $\displaystyle 9:25$ .
The perimeter of the smaller one is $\displaystyle 18$ .
Find the perimeter of the larger one.

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Exercise 5

A photo is $\displaystyle 10$ cm wide and $\displaystyle 15$ cm high.
It is enlarged to a similar photo whose area is $\displaystyle 4$ times larger.

Find the width of the enlarged photo, in cm.

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Find its perimeter, in cm.

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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 $\displaystyle P$
$\displaystyle PT^{2}$ $\displaystyle =$ $\displaystyle PA\cdot PB$