Mathematics Competition Training Class Notes Elementary ...ylmass.edu.hk/~mathsclub/MCTCN/ege.pdf · Mathematics Competition Training Class Notes Elementary Geometry 119 Elementary

Mathematics Competition Training Class Notes Elementary Geometry

119

EElleemmeennttaarryy GGeeoommeettrryy Introduction to Trigonometry

What is trigonometry?

Trigonometry is an important branch of geometry. It studies the properties of

triangles, especially right-angled triangles. This is because any triangles can always

be splitted into two right-angled triangles.

Relationship of sides and angles

In any triangle, there are six variables that we are always intersected in finding them

out: lengths of three sides and sizes of three angles of that triangle. However, in a

right-angled triangle, one angle must be a right angle, and one other angle can be

deduced from the remained one using the fact that sum of all angles in a triangle is

180°; one side can also be found using Pythagoras’ theorem (x2 + y

2 = r

2). Therefore,

the number of variable is reduced to three in the right-angled triangle case, as shown

below in red:

It would be nice if there is any relationship between these three variables. In fact,

there is. If we fix the length of side “x”, then change the value of θ, we would find

that the value of y increases as θ increases and also the same for the opposite.

Moreover, this relation between x, θ and y is unique. Hence there should be a function

that:

f(x, θ) = y (19.1)

For any given x and θ.

θ x

y r

θ1

y1 θ2

y2

y3

θ3


120

If we fix θ and change x, the value of y will also change. But in this time, the ratio of y

and x is a constant! That means (19.1) can be replaced by:

( ) yf θ

x= (19.2)

All these triangles are similar. Hence the value b⁄a is constant

This function f is called the tangent of θ. This is denoted as:

tan θ (19.3)

Some books write “tg” instead of “tan”. Please do not confuse this with the tangent of

a curve. They share the same name but are two completely different concept.

The tangent is one of the six trigonometric functions. These are all functions of θ

and produces ratio of two sides. They are called sine, cosine, tangent, cotangent,

secant and cosecant, where:

sin cos

tan cot

sec csc

y xθ θ

r r

y xθ θ

x y

r rθ θ

x y

= =

= =

= =

(19.4)

By convention, if f is a trigonometric function, we write (f(θ))n as f

n(θ) for positive n,

e.g. sin2 θ means (sin θ)

2.

Example 19a: Given the length of two sides of a triangle are a and b and their

included angle is θ. Find the area of this triangle.

Solution 19a:

θ

a

b

h

As shown in the picture, h is the altitude (“height”) to side b of the big triangle.

Sine

Tangent

Secant

Cosine

Cotangent

Cosecant


121

Therefore the area of that triangle is 12bh . But h = a sin θ. Thus the area is:

12

sinab θ (19.5)

Transforming between trigonometric functions

Using the fact that x2 + y

2 = r

2 and the definition of the trigonometric functions, one

can deduce values of all trigonometric functions from one, using any of the following

identities:

� 1

cottan

θθ

= (19.6)

� 1

seccos

θθ

= (19.7)

� 1

cscsin

θθ

= (19.8)

� sin

tancos

θθ

θ= (19.9)

� sin2 θ + cos

2 θ = 1 (19.10)

� 1 + tan2 θ = sec

2 θ (19.11)

� 1 + cot2 θ = csc

2 θ (19.12)

Example 19b: Prove (19.10)

Solution 19b:

2 22 2

2 2

2 2

2

2

2

sin cos

1

x yθ θ

r r

x y

r

r

r

+ = +

+=

=

=

(19.13)


122

Example 19c: Let θ = 45°. Find the value of the six trigonometric functions of θ.

Solution 19c:

In the figure, ABCD is a square with side length 1. Hence ∆BCD is a right-angled

triangle and ∠BDC = 45°. Thus, by the definition of tangent:

tan 45° = 1. (19.14)

And by the transformation formulae of trigonometric functions:

2

2

2

2

sin 45

cos 45

cot 45 1

sec45 2

csc45 2

° =

° =

° =

° =

° =

(19.15)

Extension of trigonometric function to [0°, 360°]

Using the definition by right-angled triangle, the domains of trigonometric functions

are limited to only [0°, 90°). Can we extend the domain to the whole real line? It is

possible, but we need to change the definition a bit using coordinates geometry.

In the figure, P (x, y) is a point on the coordinates system, r is the length of OP and θ

is the inclination of the line OP. The definitions are same as (19.4), except the

meaning of x, y, r and θ are changed. The coordinates geometric definition is

consistent with the right-angled triangle one because ∆OPQ is a right-angled triangle.

What is better than the right-angled triangle definition is that the range of θ is

extended to [0°, 360°].

When 0° < θ < 90°, all values of trigonometric functions are positive as expected. But

when 90° < θ < 180°, the x-coordinates of P will be negative. Hence the value of

A B

C D 45°

1

1

P (x, y)

O

y

x θ

r

Q (x, 0)


123

cosine, tangent, cotangent and secant will be negative using the definition. Similarly,

when 180° < θ < 270°, sine, cosine, secant and cosecant are negative and when

270° < θ < 360°, sine, tangent, cotangent and cosecant are negative.

Putting the signs of the trigonometric functions into each quadrant, we found that:

The signs of cotangent, secant and cosecant can be determined using sine, cosine and

tangent, thus they are not written. The letters in red resembles the word “CAST”. This

is called the “CAST” diagram for trigonometric functions.

Radian measure

For practical usage, the “degree” in convenient to use. However, for pure mathematics

usage, another unit called radian is more suitable. The radian is defined as:

If O is a center of a circle and A, B are points on it,

And the length of arc AB is s and radius of the circle is r,

Then radianss

AOBr

∠ = (19.16)

If AOB is a straight line, than ∠AOB = 180°. But s = πr. Therefore ∠AOB = π radians.

Thus:

180° = π radians (19.17)

We use this to converse between degree and radian measures. Usually the word

“radian” is dropped. Thus, “a right angle is π⁄2” just mean “a right angle is

π⁄2 radians”.

All > 0 Sine > 0

Tangent > 0 Cosine > 0

I II

III IV

s

r O

A

B θ


124

Example 19c: Find the area of segment AOB (shaded yellow) above if θ is measured

in radian.

Solution 19c:

If θ is measured in degree, the area can be found using:

2

360

θπr ⋅

° (19.18)

If we convert θ and 360° into radian, the result will be:

2 21

2 2

θπr r θ

π⋅ = (19.19)

By definition, s

θr= . Thus (19.18) can also be expressed as:

1

2rs (19.20)


125

Trigonometry in Depth

Transforming back to [0, /2)

The domains of trigonometric functions can actually be extended to R. This very is

nice; however, it is not efficient to calculate. Trigonometric functions are different

from polynomial or exponent that you cannot compute it using simple arithmetic.

Therefore, we should find a way to move those extended values back into )20, π ,

which is easier to work out.

Multiple revolution

When using the coordinates system to define trigonometric functions, the selection of

θ is not unique. For example, the following values of θ will all result in the same

position of P if r is fixed:

In fact, the number of revolution an angle had made does not affect the value of

trigonometric functions. Therefore, for any trigonometric functions f and integers n:

f(2nπ + θ) = f(θ) (20.1)

Negative angle

A negative angle means the angle is measured clockwise instead of counterclockwise.

It could be proved that P’ is a reflection of P along the x-axis. Hence:

( )( )( )

sin sin

cos cos

tan tan

θ θ

θ θ

θ θ

− = −

− =

− = −

(20.2)

P r

θ

2π + θ

-2π + θ

P (x, y)

r θ

-θ

P’ (x, -y)

r


126

Addition of π/2, π or 3π/2

When an angle in the first quadrant is added with π⁄2, π or

3π⁄2, it will be rotated to the

second, third or fourth quadrant.

With these special kinds of rotation, the final position can be determined easily.

Regarding to the graph above:

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )

2 2 2

3 3 32 2 2

sin sin cos cos tan tan

sin cos cos sin tan cot



π π π

π π π

θ θ θ θ θ θ

θ θ θ θ θ θ

π θ θ π θ θ π θ θ

θ θ θ θ θ θ

= = =

+ = + = − + = −

+ = − + = − + =

+ = − + = + = −

(20.3)

Notice the pattern of the result: the trigonometric function and its “co-function”

appear alternately.

Using the result of this and the two mentioned before, we can successfully transform

any θ back into )20, π without modifying the values of its trigonometric functions

significantly. These collections of formulae are called the transformation formulae

in trigonometry.

Example 20a: Given that sin 20° = a. Find tan 33550° in terms of a.

Solution 20a:

Using (20.1), tan 33550° = tan 70°.

Using (20.3) and (20.2), tan 70° = - cot (-20°) = cot 20°.

But 2 2cos 20 1 sin 20 1

cot 20sin 20 sin 20

a

a

° − ° −° = = =

° °.

Therefore 21

tan33550a

a

−° =

θ

P1 (x, y)

P2 (-y, x)

π/2 + θ

P3 (-x, -y)

π + θ

P4 (y, -x)

3π/2 + θ


127

Compound angle formulae

Frequently we would encounter expressions like sin (a + b) or cos (a – b) in

trigonometric problems. In fact, we can express these in terms of individual

trigonometric functions like sin a, cos b, etc.

Example 20b: Express sin (a + b) in terms of sin a, sin b, cos a and cos b.

Solution 20b:

x

y

h

ba

As figured, the area of red triangle is 12

sinxh a , the blue triangle is 12

sinhy b and sum

of area of the two triangles is ( )12

sinxy a b+ . Thus:

( )sin sin sinh h

a b a by x

+ = + (20.3)

But cos , cosh h

b ay x= = . Therefore:

sin (a + b) = sin a cos b + sin b cos a (20.4)

(20.4) is true for any values of a and b. Thus, if we replace b with –b, we’ve got:

sin (a – b) = sin a cos b – sin b cos a (20.5)

Using (20.2) and (20.3) to transform it into cosine, we have:

cos (a + b) = cos a cos b – sin a sin b

cos (a – b) = cos a cos b + sin a sin b (20.6)

Dividing (20.4) or (20.5) by (20.6) or reverse we have:

( )

( )

tan tantan

1 tan tan

cot cot 1cot

cot cot

a ba b

a b

a ba b

a b

±± =

± =±

∓

∓ (20.7)

(20.4) to (20.7) are called the compound angle formulae. Many identities of

trigonometric functions can be derived from this, such as:


128

Double angle formulae

These are the formulae that evaluate f(2a) for trigonometric functions f. This can be

done by setting b = a in the compound angle formulae. The results are:

2 2 2 2

2

2

sin 2 2sin cos

cos 2 cos sin 2cos 1 1 2sin

2 tantan 2

1 tan

cot 1cot 2

2cot

a a a

a a a a a

aa

a

aa

a

=

= − = − = −

=−

−=

(20.8)

Half angle formulae

These are the formulae that evaluate f(a⁄2). These can be obtained by solving (20.8).

The results are:

1 cossin

2 2

1 coscos

2 2

a a

a a

− = ±

+ = ±

(20.9)

The sign is taken according to the “CAST” diagram.

Subsidiary angle form*

This is the name of an alternate representation of a sin θ + b cos θ. That form looks

like r cos (θ – α). The value of r and α can be found using (20.6). The result is:

2 2 2

tan

r a b

aα

b

= +

=

(20.10)

Sum-to-product and product-to-sum formulae

It is possible to express the sum or difference of two identical trigonometric functions

as product of two trigonometric functions and vice-versa. Directly from the compound

angle formulae,

( ) ( )( ) ( )( ) ( )( ) ( )

sin sin 2sin cos

sin sin 2cos sin

cos cos 2cos sin

cos cos 2sin sin

a b a b a b

a b a b a b

a b a b a b

a b a b a b

+ + − =

+ − − =

+ + − = + − − = −

(20.11)

Therefore:


129

( ) ( )( )( ) ( )( )( ) ( )( )( ) ( )( )

12

12

12

12

sin cos sin sin

cos sin sin sin

cos cos cos cos

sin sin cos cos

a b a b a b

a b a b a b

a b a b a b

a b a b a b

= + + −

= + − −

= + + −

= − + − −

(20.12)

These are called the product-to-sum formulae. If we set ,2 2

x y x ya b

+ −= = , then:

2 2

2 2

2 2

2 2

sin sin 2sin cos

sin sin 2cos sin

cos cos 2cos cos

cos cos 2sin sin

x y x y

x y x y

x y x y

x y x y

x y

x y

x y

x y

+ −

+ −

+ −

+ −

+ =

− =

+ =

− = −

(20.13)

These are called the sum-to-product formulae.

Example 20b: Given 1

sin6 2

π= . Find

5sin12

π.

Solution 20b:

Let’s demonstrate two methods to solve this problem.

Method I: Direct use of compound angle formulae

Notice that 5

12 4 6

π π π= + . Therefore:

2

5sin sin cos sin cos12 6 4 4 6

1 2 2 11

2 2 2 2

2 6

4 4

2 6

4

π π π π π= +

= ⋅ + ⋅ −

= +

+=

(20.14)


130

Method II: Using half angle formulae

( )

56

6

212

3

2

5 1 5sin sin12 2 6

1 cos 5 is in first quadrant

2 12

1 cos

2

1 1

2

1

2

2 3

2

π

π

π π

π

= ⋅

− =

+=

+ −=

+=

+=

(20.15)

Note that the two methods result in answers that have totally different form, even

though their numerical values are the same (do you know how to prove?). Usually

using compound angle formulae is preferred to half angle formulae, because square

roots are harder to calculate or manipulate. But if you don’t know the values of

trigonometric functions of the two angles , you have to stick with half angle formulae.


131

Graphs of trigonometric functions*

Since trigonometric functions are f: R → R, we can draw a graph y = f(x) on a

coordinates system to investigate their properties.

x

y

−0.5π 0 0.5π π 1.5π 2π 2.5π 3π 3.5π

-1

1

y = sin x

y = cos x

The graphs of curves y = sin x and y = cos x. Note that x are in radian.

x

y

−0.5π 0 0.5π π 1.5π 2π 2.5π 3π 3.5π

-4

-2

2

4y = tan x

y = cot x

The graphs y = tan x and y = cot x.

x

y

−0.5π 0 0.5π π 1.5π 2π 2.5π 3π 3.5π

-4

-2

2

4

y = csc x

y = sec x

The graphs y = sec x and y = csc x.


132

From the graphs, we know the followings:

� All trigonometric functions are not injective.

� All trigonometric functions except tangent and cotangent are not surjective.

� ∀x ∈ R: -1 ≤ sin x, cos x ≤ 1.

The graph of y = sin x is particularly interesting. It resembles the waveform of all

ordinary waves. Therefore we could name some variables of the graph using physical

terms:

x

y

0 0.5π π 1.5π 2π 2.5π

-2

2y = a sin(bx + c)

In this figure, a = 3, b = 2, c = π/3

In physics, phase is the value of the “angle” that the first equilibrium position (zero)

is reached. Changing the phase in the same as translating the curve horizontally. The

amplitude is the maximum displacement of the wave from its equilibrium position.

Changing it means scaling the curve in the y-direction. The period is the minimum

length that a wave repeats itself. Changing this is the same as scaling the curve in the

x-direction.

Amplitude = a Phase = -c

Period = 2π/b


133

Lines and Line Segments

Definition of lines and line segments

Just like the definition in coordinates geometry, line segment is the shortest path from

one point to another point. Line is an extension of line segment to infinite.

Line and line segment

However, in geometry the term “line” usually includes “line segment” as well. In this

book, we will use the term “line” if no confusion is made.

Angles of a line

If we draw a line AB, then put a point O between A and B, then ∠AOB must be 180°.

This is derived from definition. If we break down ∠AOB using several lines passing

through O into smaller angles, then the sum of these angles are also 180°.

∠AOC + ∠COD + ∠DOE + ∠EOF + ∠FOB = 180°

This theorem is called “sum of adjacent angles on a straight line is two right

angles”, abbreviated “adj. ∠s on st. line”.

We can extend this to the other side of the line. The result is that sum of all angles

around the point O is 360°. This is true even there isn’t a “straight” line AOB.

∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOF + ∠FOA = 360°

This theorem is called “sum of adjacent angles at a point is four right angles”,

abbreviated “adj. ∠s at pt.”.

A O B

C D E F

C

D E

F

B A

O


134

Example 21a: Prove that ∠AOB = ∠COD:

Solution 21a:

( )

( ) ( )

180

180 180

AOB AOC

COD

COD

∠

∠

∠ = ° −∠

= ° − ° −∠

= ∠

adj. s on st. line

adj. s on st. line (21.1)

This theorem is called “vertical opposite angles are equal” or “vert. opp. ∠s”.

Parallel lines

Two lines are parallel if and only if they do not intersect even extended. We mark a

pair of parallel lines by adding arrow signs on them:

These two lines are parallel

When two lines AB and CD are parallel, we write

AB // CD. If another line crosses a pair of parallel

lines, then some beautiful equalities about angles

appears. In the figure, it can be proved that:

� ∠EFB = ∠EGD (21.2)

� ∠AFH = ∠EGD (21.3)

� ∠BFG + ∠FGD = 180° (21.4)

These theorems are called “corresponding angles of parallel lines are equal”

(corr. ∠s, AB // CD), “alternate angles of parallel lines are equal” (alt. ∠s, AB// CD)

and “interior angles of parallel lines are supplementary” (int. ∠s, AB // CD)

respectively.

(Supplementary means the sum of two angles is exactly two right rights, i.e., 180°)

The converses of these theorems are also true, i.e., if any of (21.2) to (21.4) is true,

then AB // CD. But the abbreviations are changed to “corr. ∠s eq.”, “alt. ∠s eq.” and

“int. ∠s supp.” correspondingly.

A

B

O

C

D

G

F

A

BC

E

H

D


135

Example 21b: Prove (21.2)

Solution 21b:

Using coordinates system, assume the slope of AB = slope of CD = m and

slope of EH = k. Since the tangent of the angle between two lines is solely depend of

the slope of the two lines, and slope of AB and CD are the same, the angle between

AB and EH and CD and EH are the same (if they are in the same quadrant). Q.E.D.

Example 21c: Prove that the sum of all interior angles of a triangle is two right

angles (∠ sum of ∆).

Solution 21c:

As figured, ∆ABC is a triangle, PQ is a

line passing through A that is parallel to

BC. Therefore:

( )( )

//

//

PQ BC

PQ BC

BAP ABC

QAC ACB

∠

∠

∠ = ∠

∠ = ∠

alt. s,

alt. s, (21.5)

But:

∠APB + ∠BAC + ∠CAQ = two right angles (adj. ∠s on st. line) (21.6)

Hence the original statement is true. Q.E.D.

Intercepts

In geometry, intercept is a line segment of a line which is cut out by two other lines.

For example, the segment PQ is an intercept made by AB and CD on XY.

If four intercepts on a two lines AE and BF are made by three distinct lines AB, CD

and EF, then the intercept theorem states that:

// //AC BD

AB CD EFCE DF

⇒ = (21.6)

BC

A P

Q

A

B

C

D

P QX

Y

A B

C D

E F


136

The converse of intercept theorem states that:

( ), // //AC BD CE DF AB CD EF= = ⇒ (21.7)

An important case is that when A coincides with B. ∆AEF will be a triangle. The

intercept theorem will be named equal ratios theorem instead:

//AC BD

CD EFCE DF

⇔ = (21.8)

The equal ratios theorem will become midpoint theorem if C is the midpoint of AE

and D is the midpoint of AF:

12

//CD EFAC CE

CD EFAD DF

= ⇒

== (21.9)

A

C D

E F


137

Triangles

What is a triangle?*

In geometry, a triangle is the region totally enclosed by three line segments.

The triangle ∆ABC

The three line segments are called the sides on a triangle. The intersections of these

lines are the vertices of the triangle. We name a triangle according to its vertices. If

the vertices are called A, B and C, then the triangle is called ∆ABC.

Angles of triangle

To describe an interior angle at the vertex K in a triangle, we usually

just write ∠K for simplification.

For any triangle ∆ABC, we have shown that:

∠A + ∠B + ∠C = 180° (22.1)

If we extend BC to any point E outside the triangle, then ∠ACE = 180° – ∠C. This

angle is called the exterior angle of a triangle at C. It immediately follows that:

∠ACE = ∠A + ∠B (22.2)

This theorem is called “exterior angle of triangle is the

sum of the two interior angles not next to it”

(ext. ∠ of ∆)

Example 22a: An n-sided polygon, or simply “n-gon” is defined by the interior of n

line segments that connects n points (vertices) on a plane.

A polygon

a

b

c

C

B

A

K

∠K

A

B C

E


138

Find the following in terms of n:

� Sum of all interior angles.

� Sum of all exterior angles.

Solution 22a:

Any polygon can be view as a combination of n triangles, like:

Assume all the triangles share a common vertex P. Sum of all interiors angles of all

triangles is 180°n. It is the same as the sum of all interior angles of the n-gon, except

the extra angles around P, which are not part of the interior angles. To encounter for

this, subtract the value by 360°. Therefore, the angle sum is given by:

180° (n – 2) (22.3)

This is called “sum of all interior angles of polygon is (n – 2)π” (∠ sum of polygon).

Since exterior angle and interiors angle are supplementary, we have:

( )

( ) ( )

1 1

1 1

exterior angles 180 interior angles

180 interior angles

180 180 2

360

n n

k k

k k

n n

k

k k

n n

= =

= =

∠

= ° −

= ° −

= ° − ° −

= °

∑ ∑

∑ ∑ sum of polygon

(22.4)

Therefore, we have “sum of all exterior angles of polygon is four right angles”

(ext. ∠ sum of polygon).

Congruence and similarity

In geometry, there are two very important relationships: congruence and similarity.

Formally, if two figures can overlap using translation, reflection and rotation only,

they are congruent. If scaling must be used, they are similar. Informally, if two figures

have the same shape and size, they are congruent. If they have same shape but

different size, they are similar.

The figure on the right is congruent to the middle one, and similar to the left one.

P


139

If two figures are congruent, then all their corresponding measures (length, area, angle,

etc.) are equal. The converse is also true.

If two figures, say figure A and figure B, are similar, and figure A contains an

n-dimensional measure A1 and m-dimensional measure A2, and similar for figure B,

then:

1 2

1 2

n m

A A

B B

=

(22.5)

Dimensions Measures

0 Angle, Ratio

1 Length

2 Area

3 Volume

The zero-dimensional measures of two similar figures are equal.

If figure A is congruent to figure B, we write A ≅ B. If A is similar to B, we write

A ~ B.

The above information is available for any figures. This is of course also true for

triangles. If fact, if ∆ABC ≅ ∆XYZ, then:

( )( )

, ,

, ,

a x b y c z

A X B Y C Z

≅

∠ ≅

= = =∠ = ∠ ∠ = ∠ ∠ = ∠

corr. sides, ∆s

corr. s, ∆s (22.6)

If ∆ABC ~ ∆XYZ, then:

( )

( ), ,

a b c

x y z

A X B Y C Z ∠

= =∠ = ∠ ∠ = ∠ ∠ = ∠

corr. sides, ∆s

corr. s, ∆s

∼

∼

(22.7)

These can be used to solve geometric problems easily. But how could we find two

congruent / similar triangles and prove them to be?

The prerequisite for a pair of triangles to be congruent is that their corresponding

angles, sides and area are all equal according to the theorem concerning congruence.

A

B C

X

Y

Z

a

b c

x

y

z


140

C2

C1

A B

But from (19.5) the area of a triangle can be found using two sides and an angle.

Therefore this can be eliminated from the list. The rest is three sides and three angles.

We will investigate what equality should be formed so that the pair of triangles are

congruent.

If the three sides of a triangle are fixed, then so is

the triangle. To prove, consider the coordinates

system. Let A be (0, 0) and B be (c, 0). Let Γ1 be

the locus that AC is a constant and Γ2 be

the locus that BC is a constant. Then Γ1 and Γ2

are two circles. They must have two

intersections. If they have only one intersection, this

must lie on the x-axis and hence ABC is a line

segment, not a triangle. If they have no intersections, no triangles can be drawn at all.

Let’s call the two intersections C1 and C2. Then C2 is just the reflection of C1 along

the x-axis, i.e., ∆ABC1 ≅ ∆ABC2. Therefore, this triangle is fixed.

This means a unique triangle can be generated by the three sides. If we create a

function that maps R3 to the set of all triangles up to translation, reflection and

rotation, this function must be bijective. Therefore:

If the three corresponding sides of two triangle are equal,

They are congruent. (SSS) (22.8)

We can also show that a triangle is fixed if either of the following is satisfied using the

locus method:

� Two sides and their included angle are fixed. (SAS)

� Two angles and their included side are fixed. (ASA)

� Two angles and a non-included side are fixed. (AAS)

� Two sides are given and one of the non-included angle is right (RHS)

Therefore these are also the criteria for whether two triangles are congruent or not.

Congruent triangles. Since AAS can be derived from ASA, it is not listed here.

SSS SAS

ASA

RHS


141

Similar triangles are trickier. We use the locus method again. Assume the angles of a

triangle are fixed, then the following “locus” (the red lines) should appear:

A red line will intersect the blue line at B and the x-axis at C. For any ∆ABC defined

by one red line, we can scale it to another ∆ABC defined by another red line.

Therefore these two triangles are similar. Therefore all corresponding angles equal is a

condition of similarity of triangle. There are totally three different conditions:

� All corresponding angles are equal (AAA)

� All ratios of the corresponding sides are equal (3 sides prop.)

� Two ratios of the corresponding sides and their included angle are equal

(2 sides ratio, inc. ∠)

Similar triangles

Special triangles

When two sides of a triangle are equal, this is called an isosceles triangle. If all three

sides are equal then this is an equilateral triangle. Many elegant theorems are

associated with these two special triangles. If ∆ABC is isosceles (AB = AC), and M is

the midpoint of BC, then:

AM ⊥ BC (22.9)

The converse of (22.9) is also true, i.e., if AM ⊥ BC and M is the midpoint of BC then

∆ABC must be isosceles. In fact, AM bisects ∠BAC too.

An isosceles triangle

A

3 sides

prop.

2 sides

ratio, inc. ∠ AAA

B C M

A


142

(22.9) and its converse and be proved easily using the fact that ∆ABM ≅ ∆ACM (Using

SSS). This also leads to another important theorem “base angles of an isosceles

triangle are equal” (base ∠, isos. ∆):

∠B = ∠C (22.10)

Its converse also holds. This is called opposite sides of two angles are equal if they

are equal (sides opp. eq. ∠s):

AC = BC (22.11)

Pythagoras’ theorem*

“Geometry has two great treasures: one is the Theorem of Pythagoras; the other, the division

of a line into extreme and mean ratio [golden ratio]. The first we may compare to a measure of

gold; the second we may name a precious jewel.” – Johannes Kepler (1571 – 1630)

Pythagoras’ theorem (Pyth. thm.) is probably one of the most well-known theorems

in mathematics. It states that, for any right-angled triangles (∠C is the right angle),

a2 + b

2 = c

2 (22.5)

The converse is true. That means if (22.5) is satisfied, the triangle must be

right-angled.

Triangle inequality*

Even we are given lengths of three sides, it is not necessary we can construct a

triangle. If one side is too long that exceeds the sum of the other two sides, then no

triangles can be drawn. This is described by the triangle inequality. If three lengths a,

b and c are given, then the following must be satisfied in order to make a triangle:

a b c

b c a

c a b

< +< +

< +

(22.6)

??? 32

7

Violation of triangle inequality


143

Circles

What are circles?

Circle is the locus traced by a point moving around a fixed point called center at a

fixed distance.

The region that is enclosed by a circle is a disc. However, the term “circle” is usually

mistakenly also includes the disc. To specify the curve “circle”, “circumference” is

used.

In a circle, there are many segments, curves and region that deserve special names.

In a circle, if a segment connects two points on the circumference, this is called a

chord. For example, the lines PQ and XY on the left of the figures are chords. If a

chord passed through the center like XY, this is called a diameter. A segment

connecting the center and a point on the circumference (e.g., OX, OY and OZ) is a

radius.

The region that is enclosed by a chord and the circumference is called a segment. The

green area is one example. However, there are two types of segment: the smaller one

like VWU that do not include the center, and the larger one like VSRTU that includes

the center. The smaller one is called the minor segment, and the larger one is the

major segment. If not specified, the minor segment is taken. The region enclosed by

two radii and the circumference like the yellow area is called a sector. Again there are

minor sector and major sector, and the minor one is used.

A section of a circumference is called an arc. If an arc connects two point A and B as

figured, we write this as �AB . The minor arc is mentioned. To mention the major

D

A

B

C

O Y O

P Q

X

Z

S

O T

V

U

W

R


144

arc connecting A and B, write �ACB instead so to specify the arc passes through the

point C also. If an arc is exactly half the length of the circumference likes �CD , we

call this a semicircle.

Special circles

A circle can be defined by a center and a radius. If two circles share the same value of

any of these, they have a special relationship. If the two circles are equal in radius,

they are called a pair of equal circles, as they are congruent to each other. If they have

the same center, they are concentric circles.

Equal circles and concentric circles

Besides, a circle can also be defined by a polygon. If a circle falls inside a polygon

and tangents to every sides of it, this is called an inscribed circle or incircle of that

polygon. If all vertices of the polygon are on the circle, this is called the

circumscribed circle or circumcircle of the polygon.

The incircle and circumcircle of a triangle

Chords of circle

As shown on the right, AB and CD are chords of circle

O. M and N are points on these chords. If OM ⊥ AB, it

can easily be shown that ∆OAM ≅ ∆OBM (RHS).

Therefore M is the midpoint of AB. The converse is

also true. Thus we have the following theorem:

N

M

O

A

B

C

D


145

D

O

A

B

C

AM = BM ⇔ OM ⊥ AB (23.1)

Moreover, if we assume OM = ON, then using ∆OAM ≅ ∆OCN (RHS) and (23.1) we

have AB = CD. Again its converse is true, thus:

OM = ON ⇔ AB = CD (23.2)

Example 23a: In the figure, ∆ABC is an isosceles triangle with

AB = AC. O is the center of the circumcircle of ∆ABC. D is a

point which the lines AO and BC intersect. If AB = 8 and the

sum of the distances from O to AB and AC respectively is 6,

find OD.

Solution 23a:

Let M and N be the feet of perpendicular of AB and AC from O

respectively. Then from (23.2), OM = ON = 3, and from (23.1), AM = 4. Using

Pythagoras’ theorem, the radius of circle O = OA = OB = OC = 5. And since D is

midpoint of BC, using (23.1) we have OD ⊥ BC. From the definition of trigonometric

functions, 35

cos cosMOB MOA∠ = = ∠ . Hence:

( )

( )

2

235

725

cos cos 2

cos 2

1 2cos

1 2

BOD π MOB

MOB

MOB

∠ = − ∠

= − ∠

= − ∠

= −

=

(23.3)

Therefore 7 725 5

OD OB= ⋅ = .

Angles of circle

Two kind of angles can be drawn specially in a circle. One is the angle between two

radii ∠AOB, which is called the angle at the center; the other is the angle between

two chords ∠APB, which is called angle at the circumference.

A

B

O

P


146

Example 23b: Prove that angle at the center is twice angle at the circumference

(∠ at center twice ∠ at ⊙ce)

Solution 23b:

Q

O

B A

P

As shown, ∆OAP is an isosceles triangle. Thus ∠OAP = ∠OPA (base ∠s, isos. ∆).

Therefore ∠AOQ = 2∠APO (ext. ∠ of ∆). This is similar for the right hand side.

Combining, we have:

∠AOB = 2 ∠APB (23.4)

This is equivalent to the original statement. Reader should also try to prove the two

cases below:

O A

A

O

P

B

P

B

From example 23b we have another important theorem:

Angles in the same segment are the same (∠s in same segment)

If the angle at center is 180° (i.e., AB is a diameter), then the angle at circumference

must be a right angle:

Angle in a semicircle is right (∠ in semicircle)

DO

A B

P

Q

C

R


147

Relationships among different parts of circles

C

D

O

B

A

In a circle, there are many different parts. How they are related to each other? First of

all, recall that an angle is defined as the ratio of arc length and radius in radian

measure. Here in the same circle, the radii are the same. Hence:

( ) � �( )AOB COD AB CD∠ = ∠ ⇔ = (23.5)

Moreover, the chords AB and CD are also equal as ∆AOB ≅ ∆COD:

( ) ( )AOB COD AB CD∠ = ∠ ⇔ = (23.6)

The areas of the two triangles are also the same. The areas of the sectors are also the

same using (19.20). So the areas of the sections are also the same. Therefore the

following statements are mutually equivalent:

� Angle at center / circumference are equal.

� Arc lengths are equal.

� Chords are equal.

� Areas of sectors are equal.

� Areas of sections are equal.

These theorems apply to equal circles as well. However, if one measure, say, angle,

does not satisfy the equality, what will happen to the others? From the experience of

defining the radian, we know that:

�

�

Area of sector

Area of sector

AOB AB AOB

COD CODCD

∠= =

∠ (23.7)

Simply because the radii are equal. But the ratio does not hold for chords, for

example:


148

2 r

r

r

r

r

r

60°

Q

R

SOP

In the figure, ∠POQ : ∠ROS = 4 : 3, but : 2 :1PQ RS = .

Example 23c: (HKCEE 1994/2) In the figure, OABCD is a sector of a circle. If

� � �AB BC CD= = , find the value of x.

x

P

C

B

A

O D

Solution 23c:

If we know ∠CAO and ∠BDO, we can find x readily applying “∠ sum of quad.” in

OAPD.

Now since the three arcs are equal, we have ∠AOB = ∠BOC = ∠COD = 30°.

Therefore ∠BOD = 60°. Using “∠ sum of ∆” and “base ∠s, isos. ∆”, we have

∠BDO = 60°. Similarly, ∠CAO = 60°. Therefore x = 360° – 60° – 60° – 90° = 150°.

Tangents of circle

Tangent of a circle is a line that passes through the

circle at only one point. If a line is tangent to a circle

at T, then the radius OT must perpendicular to it

(tangent ⊥ radius). The converse of it holds.

Generally, if a point is given outside a circle, then two

tangents can be drawn passing that point.

O T


149

U

T

O

P

T

A

PQ

B

Q

P

C

T

B

A

In the figure, PT and PU are two tangents drawn

from P to circle O. Thus ∆PTO ≅ ∆PUO (RHS).

Hence we have:

� PT = PU (23.8)

� ∠POT = ∠POU (23.9)

� ∠OPT = ∠OPU. (23.10)

These theorems are collectively called the

tangent property.

Sometimes the line TU is called the

polar of P with respect

to circle O. The point P is also

called the pole of line TU.

If a tangent and a chord of a circle intersect

at the circumference, the angle between

them is called a tangent-chord angle.

∠ATP and ∠ATQ are a pair of

tangent-chord angle. Each angle lies inside a

segment divided by the chord. The angle

at the circumference at the segment

opposite to the tangent-chord angle is called an angle at the alternate segment. For

example, ∠ABT is the angle at the alternate segment of ∠ATP. It can be shown that

the tangent chord angle and the angle at the alternate segment are always the

same (∠ in alt. segment). The converse is also true.

Example 23d: Prove ∠ in alt. segment.

Solution 23d:

As figured, we need to prove ∠PTA = ∠TBA. Now construct a

line CT such that CT ⊥ PQ. Hence CT is the diameter

(tangent ⊥ radius). Therefore ∠CAT is a right angle

(∠ in semicircle). Moreover, ∠ACT = ∠ABT

(∠s in same segment). So ∠ATC = 90° – ∠ABT and

∠PTA = 90° – ∠ATC = ∠ABT.

Q.E.D.


150

Example 23e: (HKMO 1991/HG Q10) As figured, two chords AB and CD intersect at

O. The tangents of the circle at A and C intersect at X, and the tangents at B and D

intersect at Y. If ∠AXC = 130°, ∠AOD = 120° and ∠BYD = k°, find k.

k°

130°

120°

Y

X

O

A

B

D

C

Solution 23e:

By the tangent property, we know that ∆AXC is an isosceles triangle. Therefore,

∠XCA = ∠XAC = 25°. Thus ∠ADC = 25° (∠ in alt. segment). In ∆AOD, we have

∠DAO = 35°. Apply “∠ in alt. segment” again to get ∠DBY = 35°. But ∆BYD is also

isosceles (tangent property), so ∠BDY = 35°. Hence ∠BYD = 110°, which means k is

110.


151

Quadrilaterals

What is quadrilateral?

Any polygon with four non-coincide sides is considered as a quadrilateral. There are

three classes of quadrilaterals, known as convex quadrilateral, concave

quadrilateral and skew quadrilateral.

From the left: Convex, concave and skew quadrilateral.

The convex quadrilateral is the most commonly encountered one in geometry. It can

be further classified as cyclic quadrilateral, tangential quadrilateral, trapezoid,

parallelogram, etc, which will be discussed later in this chapter.

Basic facts of quadrilateral

As discussed in example 22a, the sum of all interior angles of a quadrilateral is always

360° (∠ sum of quad.). The sum of all exterior angles is also 360°. This is not true for

skew quadrilateral, however, that the sum is (normally) unpredictable.

A quadrilateral has exactly two diagonals connecting the opposite vertices. The

diagonals together divide a convex quadrilateral into four regions.

Z

Y

X

W

E

A

D

C

B

It is true that the product of two opposite regions is invariant, i.e.,

WY = XZ (24.1)

Example 24a: Prove (24.1)

Solution 24a:

Using the notation in that figure, let AE = α, BE = β, CE = γ, DE = δ and ∠AEB = θ.

Hence 1 1 1 12 2 2 2

sin , sin , sin , sin ,αβ θ βγ θ γδ θ δα θ= = = =W X Y Z and directly

multiplying gives the desired equality. Q.E.D.

(If you are confused, recall that sin (π – x) = sin x.)


152

Trapezoids

Trapezoids (also called trapezium in British) are quadrilateral that one pair of

opposite sides are parallel.

Trapezoid

The necessary and sufficient condition for a quadrilateral being a trapezoid is that

the a pair of adjacent angles is supplementary, e.g.,

∠B + ∠C = 180° (24.2)

This is a direct result of int. ∠s, AB // CD (Assume the parallel sides are AB and CD).

Note: P is the necessary condition for Q means P ⇒ Q. P is the sufficient condition

of Q means Q ⇒ P.

Another necessary and sufficient condition is that the intersection of the diagonals

divides them into equal ratios, i.e.,

AE BE

EC ED= (24.3)

This can be shown using ∆ABE ~ ∆CDE (AAA) (So we know the value of the ratio too).

This leads to a theorem concerning the area of the four “regions”:

= =X Z WY (24.4)

A trapezoid is called an isosceles trapezoid when the two non-parallel sides are equal

in length. The “base angles” are equal in an isosceles trapezoid.

Isosceles trapezoid

Parallelogram

A parallelogram is special type of trapezoid such that both pairs of opposite sides are

parallel.

The properties of parallelogram includes:

� Opposite angles are equal (opp. ∠s, // gram), i.e.,

∠A = ∠C and ∠B = ∠D (24.5)


153

� Opposite sides are equal (opp. sides, // gram), i.e.,

AB = CD and BC = DA (24.6)

� Diagonals bisect each other (diags, // gram), i.e.,

AE = CE and BE = DE (24.7)

� One diagonal divides two equal area (diag bisects area), i.e.,

(W + X = Y + Z) and (X + Y = W + Z) (24.8)

� The area of the four “regions” are equal, i.e.,

W = X = Y = Z (24.9)

These are also necessary conditions for ABCD being a parallelogram. Moreover, if

two sides of a quadrilateral are equal and parallel, then this must be a parallelogram

(2 sides eq and //).

Rectangles, rhombuses and squares

Rectangles, rhombuses and squares are all special types of parallelograms. A

rectangle is a parallelogram that all angles are equal and right. A rhombus is a

parallelogram that all sides are equal. A square is a combination of rectangle and

rhombus.

From the left: Rectangle, rhombus and square.

These three figures have all the properties of parallelogram, in additional to:

� For a rectangle (Rectangle properties):

� The diagonals are equal. (AC = BD)

� AE = BE = CE = DE.

� For a rhombus (Rhombus properties):

� The diagonals intersect at right angle.

� The diagonals are angle bisectors of an interior angle.


154

Example 24b: In the figure, ABCDE is a regular pentagon and AD, BE intersect at F.

Prove the BCDF is a rhombus.

F

E

DC

B

A

Solution 24b:

If BCDF is a parallelogram and BC = CD, then BCDF must be a rhombus. The latter

condition is true because ABCDE is regular. The rest is to prove BCDF is a

parallelogram.

It is known that ∠BCD = ∠BAE = ∠DEA = 108°. Moreover, ∆ABE and ∆ADE are

both isosceles triangles. Hence ∠FAE = ∠FEA = 36°. Thus ∠AFE = 108° = ∠BFD.

Since ∠BCD = ∠BFD, BCDF is a parallelogram (opp. ∠s eq.). Q.E.D.

Cyclic quadrilateral

A cyclic quadrilateral is a quadrilateral that can be inscribed in a circle. ABCD is a

cyclic quadrilateral can also be written as “A, B, C, D are concyclic” (Concyclic

means the points lie on the same circle).

F

D

A

B

C

A cyclic quadrilateral ABCD

The opposite angles of cyclic quadrilateral are supplementary (opp. ∠s, cyclic

quad.):

∠A + ∠C = ∠B + ∠D = 180° (24.10)

This can be easily shown using “∠ at center twice ∠ at ⊙ce”. It immediately derives

that the exterior angle is equal to the interior at the opposite in cyclic

quadrilateral. (ext. ∠, cyclic quad.):

∠B = ∠ADF (24.11)


155

The converses of these theorems are also true. Combining with the converse of “∠s in

same segment” (i.e., ∠CAD = ∠CBD ⇔ ABCD is cyclic), we have three ways to prove

a quadrilateral is cyclic.

Example 24c: In the figure, P, Q, R and S are on a red circle and TU, UV, VW and

WT are tangents to the circle at these points. O is the center of the red circle. Count

the number of cyclic quadrilaterals defined by any of the nine points O to W,

assuming TUVW is not cyclic.

S

U

V

W

T

O

R

P

Q

Solution 24c:

Obviously PQRS is one of them. Since OP ⊥ TU, OQ ⊥ UV, OR ⊥ VW and OS ⊥ WT,

there are 4 more cyclic quadrilaterals: OPUQ, OQVR, ORWS and OSTP. So there are

5 cyclic quadrilaterals in total.

[Note that there will be more if TUVW is also concyclic.]

Tangential quadrilateral

A tangential quadrilateral is a quadrilateral that has an incircle. If a, b, c and d are

lengths of sides of a tangential quadrilateral in order, then by tangent property we

have:

2

a b c da c b d+ + ++ = = + (24.12)

This is also a necessary condition of tangential quadrilateral.


156

More about Triangles

Adoption of labels*

In this chapter, if not specified, the triangle is ∆ABC. The sides are a, b and c

corresponding to their opposite vertices. The value s is the semi-perimeter of a

triangle, which equals to 2

a b c+ + . The expression |∆ABC| means the area of that

triangle.

Sine rule and cosine rule

We’ve seen that the condition SSS, SAS and AAS are enough to fix a triangle. That

means these if we are given lengths of the three sides of a triangle, we can find the

interior angles, and similar for the other two conditions. One technique is to split a

triangle into two right-angled triangles and apply trigonometric functions.

Example 25a: Find c if a, b and ∠C are known.

Solution 25a:

a

b

c a c

hh

E

E

B

C A A

C

B

The treatment is difference for acute-angled and obtuse-angled ∆ABC. We will only

prove the case that the triangle is acute-angled. The other case can be proved similarly

and is left as an exercise for readers.

Case I: ∆ABC is acute.

Construct a point E on AC such that BE ⊥ AC. Denote the length of BE as h.

Therefore:

sin

cos

cos

h a C

CE a C

AE b CE b a C

=

=

= − = −

(25.1)


157

( )2 2

2 2 2 2 2

2 2

sin 2 cos cos

2 cos

c h AE

a C b ab C a C

a b ab C

= +

= + − +

= + −

Pyth. thm.

(25.2)

Equation (25.2) is known as the cosine rule (also called “cosine law”, “cosine

theorem”, “cosine formula”, etc.). It is applied when:

� Finding one side if the other two sides and its included angle are known:

c2 = a

2 + b

2 – 2ab cos C (25.3)

� Finding one angle if all three sides are known.

2 2 2

cos2

a b cC

ab

+ −= (25.4)

Rearranging (25.3) will give (25.4). If C is a right angle, then cos C = 0. (25.3) is then

reduced to Pythagoras’ theorem. Thus cosine rule can be thought as a generalized case

of Pythagoras’ theorem.

Example 25b: Find the circumradius R of a triangle if ∠C and c are known.

Solution 25b:

The circumradius of a figure is the radius of the circumcircle of a figure. Similarly,

the inradius is the radius of the incircle.

a

b

c

C'

O

A C

B

As figured, we construct a point C’ on the circumcircle so that BOC’ is a straight line.

Thus ∠C = ∠C’ (∠ in same segment), ∠BAC is right (∠ in semicircle) and BC’ = 2R.

But in ∆ABC’, we have 2sin sin

AB cR BC

C C′= = =

′. Hence:

2sin

cR

C= (25.5)


158

(25.5) is also valid for {a, A} and {b, B} as well. Combining, we have:

2sin sin sin

a b cR

A B C= = = (25.6)

This is known as the sine rule (sine law, sine theorem…). Besides finding the

circumradius, the sine rule can also be used in:

� Finding one side if two angles and one side are known.

� Finding one angle if two sides and one non-included angle are known.

Using sine rule and cosine rule, we can find all other properties (typically all sides and

angles) of a triangle with three pieces of “information” only. This process is called

solving the triangle.

Example 25c: Given b = 3, ∠A = 60° and c = 5. Solve ∆ABC.

Solution 25c:

Since ∠A is included by b and c, we apply the cosine rule to find a:

( ) ( )

2 2

2 2

2 cos

3 5 2 3 5 cos60

34 15

19

a b c bc A= + −

= + − °

= −

=

(25.7)

Now we know a, b, c and ∠A. Of course we can use cosine rule to find ∠B and ∠C,

but sine rule is easier:

sin sin

5sin 60 5 3 5 57sin

3819 2 19

83.4132

C A

c a

C

C

=

°= = =

≈ °

(25.8)

∠B can be found easily using ∠ sum of ∆:

180

36.5868

B A C= ° − −

≈ ° (25.9)


159

Usually sine rule is preferred for easily calculation when there is a choice, but

ambiguity will occur using this. Consider the following figure:

c

a a

C2

A

B

C1

The values of a, c and ∠A are known, so we can apply sine rule to the value of obtain

∠C. However, there are two possible positions of C, namely C1 and C2 in the figure,

where ∠C1 ≠ ∠C2. (To be precise, ∠C1 and ∠C2 are supplementary) Therefore sine

rule can yield two sets of solution which are completely difference. To avoid this, use

cosine rule instead.

Example 25d: Given the sides of a triangle are a, b and c. Find area of the triangle.

Solution 25d:

Using cosine rule we find 2 2 2

cos2

a b cC

ab

+ −= . Thus the area of the triangle is:

( ) ( ) ( )( )

( ) ( ) ( )

12

212

22 2 2

2 2 4 4 4 2 2 2 2 2 2116

2 2 2 2 2 2 4 4 4116

116

Area sin

1 cos

11

2 2

4 2 2 2

2 2 2

2 2 2 2

ab C

ab C

a b cab

ab

a b a b c a b b c c a

a b b c c a a b c

a b c a b c b c a c a b

a b c a b c b c a c a b

s s a s b s c

=

= −

+ −= −

= − − − − + +

= + + − − −

= + + + − + − + −

+ + + − + − + − =

= − − −

(25.10)

Where 2

a b cs + += is the semi-perimeter. This equation is known as Heron’s formula

for finding the area of a triangle from the three given sides.

Area of triangle

The simplest way to calculate the area of triangle is the formula “half base times

height” taught in primary schools. The height is usually not given and at this level we

usually employ a more advanced formula “ 12

sinab C ” introduced in example 19a. In


160

the last example we proved the Heron’s formula “ ( ) ( ) ( )s s a s b s c− − − ”. Are there

more? Definitely.

Let’s consider the triangle with an incircle on the right.

The area of the red, green and blue small triangles are 1 1 12 2 2

, andar br cr respectively. (r is the inradius)

Thus the total area of the big triangle is

( )2r a b c+ + , or simply rs.

Now consider 12

sinab C . Using sine rule,

we know that sin2

cC

R= . Substitute

this back to 1

sin2

ab C we have got

4

abc

R as another formula of area. In fact, we can apply sine rule the other way to get

2the area 2 sin sin sinR A B C= . To conclude,

( )( )( )

12

12

2

∆

sin

4

2 sin sin sin

ABC bh

ab C

s s a s b s c

rs

abc

R

R A B C

=

=

= − − −

=

=

=

(25.11)

Using (25.11), we can express R and r in terms of a, b and c. Moreover, for any

tangential polygon, its area can be expressed as rs.

Common-sided triangles theorem

Two triangles that share a common side are called a pair of common-sided triangles.

Common-sided triangles are widespread in geometric figures. For example, the

three-colored triangle on the top of this page contains 6 pairs of common-sided

triangles. There is a simple yet elegant relationship among common-sided triangles

about area and segment ratio, and many well-known theorems can be proved using

this! This relationship is described by the common-sided triangles theorem. It states

that, if ∆ABC and ∆A’BC are two triangles sharing the side BC, and AA’ intersects BC

a

c

b

r

r

r


161

C' C

A

B

(produced) at M, then:

∆

∆

ABC AM

A BC A M=

′ ′ (25.12)

M

M

M

M B C

A

A'

A'

A

C B

B

C

A

A'

A'

A

C B

Four types of common-sided triangles

Example 25e: Prove (25.12)

Solution 25e:

We will first need a lemma:

If ∆ABC and ∆ABC’ are two triangles such that B, C’ and C are

collinear, then ∆

∆

ABC BC

ABC BC=′ ′

.

Proof of lemma: Notice that ∆ABC and ∆ABC’ have the same height. Hence from

“ 12bh ” the equation is obvious!

Now back to the main problem:

∆ ∆ ∆ ∆

∆ ∆ ∆ ∆

ABC ABC AMC A MC

A BC AMC A MC A BC

BC AM MC

MC A M BC

AM

A M

′= ⋅ ⋅

′ ′ ′

= ⋅ ⋅′

=′

(25.13)

Q.E.D.

Example 25f: In ∆ABC, D, E and F are points on BC, CA and AB respectively such

that each point divides the corresponding segment into ratio 1:2 internally. X, Y and Z

are intersection of AD and BE, BE and CF and CF and AD respectively. Find

∆

∆

XYZ

ABC.

From lemma


162

Solution 25f:

Z

Y X

E

D

F

A

B C

The configuration is displayed as the above figure. To find ∆

∆

XYZ

ABC, we may solve

for ∆ ∆ ∆

1∆

XAB YBC ZCA

ABC

+ +− instead, since the three little triangles are

“common-sided” with the big triangle ∆ABC. Moreover, this configuration is

symmetric about A, B and C, i.e., switching places for A, B and C does not affect to

overall result. Thus ∆ ∆ ∆

∆ ∆ ∆

XAB YBC ZCA

ABC ABC ABC= = . We will find the ratio at the middle

in this example:

( )

1

1

1

112

27

∆ ∆

∆ ∆

∆ ∆ ∆

∆ ∆ ∆

1

2 1

YBC ABC

ABC YBC

YBA YBC YAC

YBC YBC YBC

AE AF

EC FB

−

−

−

−

=

= + +

= + +

= + +

=

(25.14)

Hence ∆ 2 2 2 1

1∆ 7 7 7 7

XYZ

ABC

= − + + =

.

We can “split” ∆ABC to a

simpler form, but not ∆YBC.


163

1

30°

D

A

B C

Angle bisector theorem

K B C

A

In the figure, AK is an internal angle bisector of ∠BAC, and K is on BC. It seems that

the ratio AB : AC is equal to KB : KC by observation. In fact, it is true. This is known

as angle bisector theorem:

AB KB

AC KC= (25.15)

The prove follows:

( )( )

1 12 2

1 12 2

sin∆

∆ sin

AB AK BACAKBKB AB

KC AKC AC AK BAC AC

⋅ ⋅ ∠= = =

⋅ ⋅ ∠ (25.16)

Example 25g: Find tan 15° without compound angle formula and its extensions.

Solution 25g:

As figured, ∆ABC is a right-angled triangle with ∠ABC = 30° and

BC = 1. Therefore:

2 3

3

3

3

sec30

tan30

AB

AC

= ° =

= ° =

BD is the angle bisector of ∠ABC.

By angle-bisector theorem,

we know that:

2 3

3

AD AB

DC BC= = (25.17)

But:

3

3AD DC AC+ = = (25.18)

So:

3

3

2 3

3

3 6 3 32 3

31 3 2 3CD

−= = = = −+ +

(25.19)

By the definition of tangent, we know that tan15 2 3CDBC

° = = − .

(25.16)


164

The four centers of triangle*

In a triangle, there are many special points called centers of triangle. It is known that

there are infinitely many centers in a triangle, but we will only investigate four of

them.

Centroid

G

B'

A'

C'

A

B C

If we create a midpoint on a side of a triangle and draw a line from this point to the

opposite vertex, we’ve constructed a median. The three medians of a triangle will

concurrent at a point G, called the centroid, center of mass or barycenter. We can

easily prove that:

2AG BG CG

A G B G C G= = =

′ ′ ′ (25.20)

Using coordinate geometry, if the A = (x1, y1), B = (x2, y2) and C = (x3, y3), then:

1 2 3 1 2 3,3 3

x x x y y yG

+ + + + =

(25.21)

In fact, if the vertices of a polygon are A1 (x1, y1), A2 (x2, y2), … An (xn, yn), then the

centroid is:

1 1

1 1,

n n

k k

k k

x yn n= =

∑ ∑ (25.22)

Circumcenter

O

C'

A'

B'

A

B C

The circumcenter is the center of the circumcircle of a triangle. Using the fact that


165

radius is perpendicular to a chord and passes through it midpoint, we conclude that

the three perpendicular bisectors of a triangle intersect at the circumcenter.

Orthocenter

H

F

E

D

A

B C

An altitude is a line that passes through a vertex and perpendicular to its opposite

side. It is also known as a “height” in primary school. The point that on altitude meets

the side is called the foot of perpendicular from the vertex. The three altitudes will

intersect one a point called orthocenter. The presence of the orthocenter will result in

many cyclic quadrilaterals, namely:

CDEH, CDAF, AFEH, BDAE, BDFH and CEBF (25.23)

By converse of ∠ in same segment and opp. ∠s supp. To prove that a point H is an

orthocenter of a triangle ∆ABC, we only need to show:

� AH ⊥ BC (25.24)

� (∠BAC and ∠BHC are supplementary) or (BH ⊥ CA) or (CH ⊥ AB) (25.25)

The last two conditions of (25.25) are the same as the definition of orthocenter.

H

A'

A

B C

To show the first is also a necessary condition, reflect A along BC to A’. Then A’BCH

is a cyclic quadrilateral. The points that satisfy both (25.24) and first condition of

(25.25) are only H and A’. Obviously A’ should be discarded since

andA C AC B C BC′ ′⊥ ⊥ except ∠ABC = ∠ACB = 45°. And since the orthocenter

also satisfies both, H must be the orthocenter.


166

Incenter

I

A

B C

The incenter is the center of the incircle of a triangle. From tangent property, it is

also the concurrency of the three internal angle bisectors. To prove that a point I is the

incircle, we may show that:

� I lies on the internal angle bisector of ∠A. (25.26)

� 12

BIC π BAC∠ = + ∠ (25.27)

We can prove these by locus method.

Actually, if we consider the concurrency of two external angle bisectors and one

internal angle bisector, then they will intersect at a point called excenter. The excircle

is a circle that tangents to all sides of a triangle and centered at an excenter. The

exradius is the radius of an excircle. There are three excenters, thus three excircles

and exradii.

I

IA

IC

IB A

C B

The three excenters and excircles of a triangle.

The yellow lines are internal angle bisectors and the red, blue and green ones are external.


167

r

dr S

Q

P

O

X

R

More about Cyclic Quadrilaterals

Intersecting chords theorem

To describe the relative position with respect to a circle, we can use the distance

between the point and the center d and compare it with the radius r, i.e. calculate d – r.

Besides this, we can use a two-dimensional measure: power of a point with respect to

the circle. The power of a point is d2 – r

2. The importance if power can be shown

below:

Suppose X is a point outside the circle and OP ⊥ OX and PX

intersect the circle at Q. Then 2 2PX d r= + . Also

2 2cos

dOQX

d r∠ =

+. Thus, using cosine rule (yes,

we are using cosine rule instead of sine rule) we find

2 2

2 2

d rQX

d r

−=

+. Multiplying PX and QX

together gives the power of X. It can also be shown that even if OP and OX is not

perpendicular and/or X is inside the circle the result is still the same. Thus, the power

of a point X with respect to a circle can also be defined as:

The power of X is

The product of PX and QX where PQ is a chord

And will pass through X when produced (26.1)

The positions of P and Q do not matter. That means the power of a point defined by

(26.1) is an invariant for different “PQ”. Therefore, if P, Q, R and S are four points on

a circle and PQ, RS intersects at X, then:

PX · QX = RX · SX. (26.2)

(26.2) is known as the intersecting chords theorem. The converse of it is also true,

i.e., if (26.2) is satisfied than PQRS is a cyclic quadrilateral. The special case is that

when R = S, RX is tangent to the circle. Thus:

PX · QX = RX2 (26.3)


168

Is a special case of intersecting chords theorem and also a determination for tangency.

Example 26a: It is known that there are two circles Γ1 and Γ2 intersect at A and B. P

is a point on AB. L1 and L2 are two lines passing through P. L1 intersects Γ1 at C and D

and L2 intersects Γ2 at E and F. Prove that CDEF is a cyclic quadrilateral.

L 1

L 2

Γ2

Γ1

F

C

B

A

P

D

E

Solution 26a:

Using intersecting chords theorem,

PC · PD = PA · PB = PE · PF (26.4)

Thus, by converse of intersecting chords theorem, CDEF is a cyclic quadrilateral.

Q.E.D.

Ptolemy’s theorem*

By now, all but one theorems in proving cyclic quadrilaterals are concerning angles.

Intersecting chords theorem deals with lengths, through, you need an extra point. The

Ptolemy’s theorem we are introducing here satisfies the requirement, but it starts out

as an inequality. If ABCD is an quadrilateral, then:

AB · CD + DA · BC ≥ AC · BD (26.5)

Equality holds if and only if ABCD is cyclic.

A

D C

B


169

Example 26b: Prove Ptolemy’s theorem.

Solution 26b:

To start with, we construct a point D’ such that ∆ACD ~ ∆ABD’. In fact, from this

construction we also have ∆ABC ~ ∆AD’D (2 sides ratio, inc. ∠). (We will see how

this construction makes sense next chapter.)

D'

B C

D

A

From the similarities, we know that:

D B CA AB CD

CA D D DA BC

′ ⋅ = ⋅ ′⋅ = ⋅

(26.6)

Applying triangle inequality on ∆BDD’, we have:

D’B + D’D ≥ BD (26.7)

Equality holds if and only if D’ lies on BD (In this case BDD’ is just a line, or a

degenerated triangle). Using (26.6) and (26.7) we have:

( )AB CD DA BC CA D B D D AC BD′ ′⋅ + ⋅ = ⋅ + ≥ ⋅ (26.8)

Condition for equality holds is the same as that of (26.7).

Example 26c: Let ∆ABC be an equilateral triangle. Let X be a point on �BC on the

circumcircle of ∆ABC. Prove that XA = XB + XC.

Solution 26c:

Apply Ptolemy’s theorem on cyclic quadrilateral ABDC. The result follows

straightforwardly. Q.E.D.


170

Geometric Transformation Geometric transformation is coordinates geometric transformation without

coordinates. That means everything in geometric transformation is the same as the

corresponding one in coordinates. Therefore we will not go in detail in the following

transformations

Translation

Suppose T is a translation map. If we know T will map point A to point B, and assume

figure F will be mapped to figure F’, we write:

( )T AB

F F ′→��

(27.1)

The properties of translation are:

� F and F’ are congruent.

� If F is a line, then F’ // F and F’ = F.

� If two points X, Y are translated to X’, Y’, then XYY’X’ will be a parallelogram.

Scaling (Homothety)

If H is a scaling map with scaling factor r and O be the fixed point, we write:

( ),H O r

F F ′→ (27.2)

The properties of scaling are:

� F and F’ are similar.

� If F is a line, then F’ // F and length of F’ = |r| × length of F.

� If F is a point, then O, F and F’ are collinear.

� If F is a group of points, then lines connecting corresponding points between F

and F’ will all pass through O.

Reflection

If F is reflected along the line L to form F’, we write:

( )S L

F F ′→ (27.3)

The properties of reflection are:


� If F is a line, then F’ = F and L is the bisector of the angle between F and F’.

Rotation

If F is rotated to F’ by an angle of α with center O, we write:

( ),R α O

F F ′→ (27.4)


171

The properties of rotation are:


� If F is a line, then F = F’ and the angle between them is α.

� If F = {A, B} and F’ = {A’, B’} where A, A’, B and B’ are points then:

� ∆OAA’ ~ ∆OBB’;

� ∆OAB ≅ ∆OA’B’.

� If α = 60°, then OA = OA’ = AA’ and OB = OB’ = BB’.

Spiral similarity

Spiral similarity is the combination of rotation and scaling, where the fixed point of

scaling and the center of rotation are the same. Suppose this is O. If F is mapped to F’

by spiral similarity, then:

� F and F’ are similar.

� If F = {A, B} and F’ = {A’, B’}:

� ∆OAB ~ ∆OA’B’.

� ∆OAA’ ~ ∆OBB’.

� The angle between AB and A’B’ equals to ∠AOA’.

In example 26b, we constructed D’ by a spiral similarity with A as center and mapping

C to B.

Applications of transformation

Here we present some examples that can be solved with transformation easily.

Example 27a: (IMOPHK 1993 Q13) Let ∆ABC be an equilateral triangle. Let O be a

point inside ∆ABC such that OA = 3, OB = 4 and OC = 5. Find the area of ∆ABC.

Solution 27a:

Notice that (3, 4, 5) is a Pythagorean triplet, i.e., 32 + 4

2 = 5

2. This is a good indication

that Pythagoras’ theorem may be helpful. Therefore making a right-angled triangle

with side lengths 3, 4 and 5 is now our target.

But how to do so? It is impossible to make this “naturally” inside the triangle since

the angle between OA and OB is very big (if your drawing is good.) However, we

may construct a point O’ such that CO’ = 4 and OO’ = 3. This follows that O’ is

actually a rotation of O about A by 60° clockwise.


172

4

3

3

5

4 3

O'

O B

A

C

Therefore, ∆AOO’ is equilateral. Hence ∠AO’C = 90° + 60° = 150°. Since

∆AOB ≅ ∆AO’C (property of rotation), we have ∠AOB = 150°. Finally, we can use

cosine rule to find out AB and then the area of ∆ABC (which is 25 3

49 + ).

Example 27b: (Euler line.) Prove that the orthocenter, centroid and circumcenter of a

triangle are collinear.

Solution 27b:

Let the triangle be ∆ABC, G be its centroid, H be its orthocenter and O be its

circumcenter. Let A’ be the midpoint of BC, and similar for B’ and C’. (The equivalent

forms of equations with B and C won’t be mentioned if implied later) We have:

( )12,∆ ∆

H GABC A B C

− ′ ′ ′→ (27.5)

This is because AG : A’G = 2 : 1 from (25.10). But O is the orthocenter of ∆A’B’C’.

This follows from OA’ ⊥ BC and B’C’ // BC (property of scaling). This will mean:

( )12,H G

H O−

→ (27.6)

Therefore O, G, H are collinear. Moreover, OG : GH = 1 : 2.

Q.E.D.

Example 27c: Consider a triangle with side lengths 4, 5 and 5.5. Let X, Y and Z be

points on three different sides. Find the minimum and maximum of the perimeter of

∆XYZ.


173

Solution 27c:

The maximum is easy, which is just the perimeter of the big triangle, i.e., 14.5. The

minimum is trickier. Usually, reflection is used to minimize sum of line segments.

5.5

5

4

Z"

Z'

C

A

B Z

X

Y

Now reflect Z to Z’ along AB and to Z” along AC. Then

XY + YZ + ZX = Z”X + XY + YZ’ ≥ Z”Z’ (27.7)

That means a local minimum will be Z”Z’ when Z”XYZ’ is a straight line. Now we

want to minimize Z”Z’. By property of reflection, we have AZ = AZ’ = AZ”, or the

circumcenter of ZZ’Z” is A. Since

( )2 2

2

Z AZ ZAZ ZAZ

CAZ BAZ

CAB

′′ ′ ′′ ′∠ = ∠ +∠

= ∠ + ∠

= ∠

property of reflection (27.8)

Which is an invariant as Z moves, the length of Z’Z” is directly proportional to AZ.

This is minimum when AZ is the altitude. The length of Z’Z” can then be calculated

(which is 237513520

).


174

Simple Solid Geometry

What is solid geometry?

Until now, the geometry we have investigated is planar (in 2D). However, the space

we are living in is 3D. Hence there should be a branch dealing with 3D figures to have

a real application of geometry. The geometry that studies the figures in 3D space is

called solid geometry.

Some basic figures

There are many different kinds of objects in 3D. Here we show the important ones.

Cubes*

Cubes (sometimes hexahedron) are probably the “easiest” figure in 3D to human

beings. A cube has 6 faces which are all squares. If the length of a square is a, then the

volume (“amount of matter” inside) of the cube is a3.

Cuboid*

Cuboid is a cube with rectangular faces. The volume of a cuboid is abc.

a

a

b

c


175

Prism*

If congruent 2D figures are stacked at a same place, a prism will appear. The volume

of a prism is Ah. The property of prism is that no matter where we cut it, the face on

top and bottom (bases) must be the same as long as the cut is “parallel” to the base.

Cube and cuboid are special kinds of prism.

Cylinder*

A cylinder is a prism with circular base. Therefore the volume of a cylinder is πr2h.

Moreover, since the curved surface of the cylinder can be “flatten” to form a

rectangle with lengths h and 2πr, the total surface area of a cylinder is 2πr(r + h).

Pyramid

A pyramid is an object that is made up of a 2D figure and a point (“vertex”) such that

the edges of the base of the pyramid meet the point via straight lines. The volume of a

h

A

h

πr2

Curved surface

A

h


176

pyramid is 13Ah . Here h is the distance between the vertex and the base.

Right pyramids

A right pyramid is a pyramid such that the foot of perpendicular from the vertex to

the base matches the centroid of it. Typically, the base is a polygon. The faces (lateral

faces) except the base are isosceles triangles. Hence the edges of the lateral faces are

of the same length (the slant height s).

Cone

Cones are pyramids with circular base. Hence the volume is 213πr h . Moreover, we

can “spread out” the curved surface to form:

The area is then:

( ) ( )122πr s πrs= (28.1)

A

h

V

s

πr2

h s

Curved

surface

s

2πr


177

Bu by Pythagoras’ theorem, r2 + h

2 = s

2. Hence the total surface area is:

( )2 2πr r r h+ + (28.2)

Sphere

Sphere is an object that every point on the surface of it is equidistant from a point

“center”. The distance is called the radius. If the radius of a sphere is r, then its

volume is:

343πr (28.3)

And the surface area is:

4πr2 (28.4)

Spheroid and Ellipsoid*

If we stretch (or compress) a sphere along one direction only, we get a spheroid. If

we do the same to another direction perpendicular to the one before, we get an

ellipsoid. The “radius” along each stretched directions are called semi-axes. If the

semi-axes of an ellipsoid are a, b and c, the volume of it is:

43πabc (28.5)

r


178

Example 28a: If the volume and surface area of a sphere are numerically equal, find

the values of them.

Solution 28a:

Let the radius be r. Then:

3 24

34

3

πr πr

r

=

= (28.6)

Therefore their values are:

( )24 3 36π π= (28.7)

Example 28b: A (regular) tetrahedron is a pyramid that every face of it is an

equilateral triangle. If the side length of the triangle is x, find the volume of the object.

Solution 28b:

A tetrahedron

A tetrahedron must have 4 faces since the base is a triangle. It is also a right pyramid.

The base area of the tetrahedron is 2 3

4

x. The height is trickier, but we could also

find this. Referring to the following diagrams, length of a median of a base is 3

2

x

by Pythagoras’ theorem. Let G be the centroid of this triangle. This point will cut the

median into lengths of 3

3

x and

3

6

x. Next, if we cut the tetrahedron in half along

a median of the base, the get the figure on the right of below:

Hence 2

2 2

3 3

xh x x= − = , and the volume of the tetrahedron is

3 2

12

x.

x x

x

t t x h

2t⁄3

t⁄3


179

Summary � Trigonometry

� In the following triangle:

� sin θ = y / r (Opposite / Hypotenuse)

� cos θ = x / r (Adjacent / Hypotenuse)

� tan θ = y / x (Opposite / Adjacent)

� cot θ = x / y.

� sec θ = x / r.

� csc θ = y / r.

� Trigonometric identities:

� 1

cottan

θθ

=

� 1

seccos

θθ

=

� 1

cscsin

θθ

=

� sin

tancos

θθ

θ=

� sin2 θ + cos

2 θ = 1

� 1 + tan2 θ = sec

2 θ

� 1 + cot2 θ = csc

2 θ

� “CAST” diagram, representing the sign of trigonometric functions with

different range of θ:

θ x

y r

All > 0 Sine > 0

Tangent > 0 Cosine > 0

I II

III IV

0°

90°

180°

270°


180

� Radian measure:

� s

θr= (in radian)

� 21 1

Area of segment 2 2

AOB r θ rs= =

� 180° = π radians

� Trigonometric functions on R: (f is any trigonometric function)

� f(2nπ + θ) = f(θ)

� sin (-θ) = -sin θ

� cos (-θ) = cos θ

� tan (-θ) = -tan θ

� ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )

2 2 2

3 3 32 2 2





π π π

π π π

θ θ θ θ θ θ

θ θ θ θ θ θ

π θ θ π θ θ π θ θ

θ θ θ θ θ θ

= = =

+ = + = − + = −

+ = − + = − + =

+ = − + = + = −

� Compound angle formulae and its extensions:

� sin (a ± b) = sin a cos b ± sin b cos a

� cos (a ± b) = cos a cos b ¡ sin a sin b

� ( ) tan tantan

1 tan tan

a ba b

a b

±± =

∓

� ( ) cot cot 1cot

cot cot

a ba b

a b± =

±∓

� sin 2a = 2 sin a cos a

� cos 2a = cos2 a – sin

2 a = 2 cos

2 a – 1 = 1 – 2 sin

2 a

� 2

2 tantan 2

1 tan

aa

a=−

�

2cot 1cot 2

2cot

aa

a

−=

� 1 cos

sin2 2

a a− = ±

� 1 cos

cos2 2

a a+ = ±

� ( )( )2 2 2

sin cos costan a

b

r a ba θ b θ r θ α

α

= ++ = − ⇔

=

s

r O

A

B θ

*


181

�

( ) ( )( )( ) ( )( )( ) ( )( )( ) ( )( )

12

12

12

12

sin cos sin sin

cos sin sin sin

cos cos cos cos

sin sin cos cos

a b a b a b

a b a b a b

a b a b a b

a b a b a b

= + + −

= + − −

= + + −

= − + − −

�

2 2

2 2

2 2

2 2

sin sin 2sin cos

sin sin 2cos sin

cos cos 2cos cos

cos cos 2sin sin

x y x y

x y x y

x y x y

x y x y

x y

x y

x y

x y

+ −

+ −

+ −

+ −

+ =

− =

+ =

− = −

� Graphs of trigonometric functions: See page 131

� All trigonometric functions are not injective.

� All trigonometric functions except tangent and cotangent are not

surjective.

� ∀x ∈ R: -1 ≤ sin x, cos x ≤ 1.

� Lines and Line Segments

� Sum of adjacent angles on a straight line is 180°. (adj. ∠s on st. line)

� Sum of adjacent angles at a point is 360°. (adj. ∠s at pt.)

� Vertical opposite angles are equal. (vert. opp. ∠s)

� For parallel lines AB and CD:

� Corresponding angles of parallel lines are equal (corr. ∠s, AB // CD)

� Alternate angles of parallel lines are equal (alt. ∠s, AB // CD)

� Interior angles of parallel lines are supplementary (int. ∠s, AB // CD)

� The converses of these three theorems are true.

� Intercept theorem: (For ACE and BDF are straight lines)

� // //AC BD

AB CD EFCE DF

⇒ =

� ( ), // //AC BD CE DF AB CD EF= = ⇒

� Equal ratios theorem: (Intercept theorem with A = B):

//AC BD

CD EFCE DF

⇔ =

� Midpoint theorem (Equal ratios theorem with C as midpoints of AE):

12

//CD EFAC CE

CD EFAD DF

= ⇒

==

*

*

*


182

� Triangle

� ∠A + ∠B + ∠C = 180°. (∠ sum of ∆)

� ∠ACE = ∠A + ∠B. (ext. ∠ of ∆) (E is on BC produced)

� For an n-gon:

� Sum of all interior angles is 180°(n – 2). (∠ sum of polygon)

� Sum of all exterior angles of polygon is 360° (ext. ∠ sum of polygon)

� Figure A ≅ Figure B iff they have the same shape and size.

� Figure A ~ Figure B iff they have the same shape but different size.

� ( )( )( )

, ,∆ ∆

, ,

a x b y c zABC XYZ

A X B Y C Z

≅

∠ ≅

= = =≅ ⇒

∠ = ∠ ∠ = ∠ ∠ = ∠

corr. sides, ∆s

corr. s, ∆s

� ( )( )

( )∆ ~ ∆

, ,

a b c

x y zABC XYZ

A X B Y C Z ∠

= =⇒ ∠ = ∠ ∠ = ∠ ∠ = ∠

corr. sides, ∆s

corr. s, ∆s

∼

∼

� Two triangles are congruent iff:

� All three corresponding sides are equal. (SSS)

� Two corresponding sides and their included angle are equal. (SAS)

� Two corresponding angles and their included side are equal. (ASA)

� Two corresponding angles and a non-included side are equal. (AAS)

� Two corresponding sides are equal and one of the non-included angle

is right (RHS)

� Two triangles are similar iff:

� All corresponding angles are equal. (AAA)

� All ratios of the corresponding sides are equal. (3 sides prop.)

� Two ratios of the corresponding sides and their included angle are

equal. (2 sides ratio, inc. ∠)

� M is midpoint of BC ⇒ (∆ABC is isosceles with AB = AC ⇔ AM ⊥ BC)

� ∆ABC is isosceles with AB = AC ⇔ ∠B = ∠C

(Forward: base ∠, isos. ∆; Converse: sides opp. eq. ∠s)

� a2 + b

2 = c

2 ⇔ ∠C = 90° (Pyth. thm.)

� Triangle inequality: In a triangle,

a b c

b c a

c a b

< +< +

< +


183

� Circles

�

� Equal circles have equal radii.

� Concentric circles have same centers.

� An incircle of polygon is tangent to every sides of the polygon.

� A circumcircle of polygon contains all vertices of the polygon on the

circumference.

� If AB is a chord of circle O with M on AB:

AM = BM ⇔ OM ⊥ AB

� If AB, CD are chords, M, N are midpoints of them:

OM = ON ⇔ AB = CD

� Angle at the center = 2 × Angle at the circumference.

(∠ at center twice ∠ at ⊙ce)

� Angles in the same segment are the same. (∠s in same segment)

� Angle in a semicircle is 90°. (∠ in semicircle)

� The followings are equivalent:

� Angle at center / circumference are equal.

� Arc lengths are equal.

� Chords are equal.

� Areas of sectors are equal.

� Areas of sections are equal.

� For A, B, C, D on circle O:

�

�

Area of sector

Area of sector

AOB AB AOB

COD CODCD

∠= =

∠

� A line is tangent to a circle at T iff the radius OT must perpendicular to it.

(tangent ⊥ radius)

D

A

B

C

O O

P Q

X

Z

S

O T

V

U

W

R

Chord

Diameter Radius

Sector

Segment Semicircle

Arc

Circumference

Center

*

*


184

� Tangent property (Let PU, PT tangent to circle O):

� PT = PU

� ∠POT = ∠POU

� ∠OPT = ∠OPU

� The tangent chord angle and the angle at the alternate segment are always

the same (∠ in alt. segment).

� Quadrilaterals

� Sum of all interior angles of a quadrilateral is 360° (∠ sum of quad.).

� If four regions of a convex quadrilateral, W, X, Y and Z are separated by

the two diagonal and is in directional order, then WY = XZ.

� If trapezoid with AB // CD and E = AC ∩ BD.

� ∠B + ∠C = 180°

� AE BE

EC ED=

� ∆ ∆ ∆ ∆BEC DEA AEB CED= = ⋅

� Parallelogram:

� ∠A = ∠C and ∠B = ∠D (opp. ∠s, // gram)

� AB = CD and BC = DA (opp. sides, // gram)

� AE = CE and BE = DE (diags, // gram)

� (W + X = Y + Z) and (X + Y = W + Z) (diag bisects area)

� W = X = Y = Z

� All five theorems are conversable.

� Two sides of quadrilateral are equal and parallel ⇒ parallelogram.

(2 sides eq and //).

� Rectangle is parallelogram that all angles are equal and right:

� Diagonals are equal. (AC = BD)

� AE = BE = CE = DE.

� Rhombus is parallelogram that all sides are equal.

� Diagonals intersect at 90°.

� Diagonals are angle bisectors of an interior angle.

� Square is combination of rectangle and rhombus.

� Cyclic quadrilateral is quadrilateral that has a circumcircle.

� ∠A + ∠C = ∠B + ∠D = 180° (opp. ∠s, cyclic quad.)

� ∠B = ∠ADF (ext. ∠, cyclic quad.) (F is on CD produced)

� Both theorems are conversable.

� Converse of ∠s in same segment is determination of cyclic quadrilateral.


185

� Tangential quadrilateral is a quadrilateral that has an incircle.

� 2

a b c da c b d+ + ++ = = +

� More about Triangles

� Cosine rule:

� c2 = a

2 + b

2 – 2ab cos C

�

2 2 2

cos2

a b cC

ab

+ −=

� Sine rule: 2sin sin sin

a b cR

A B C= = =

� Area of triangle:

( )( )( )

12

12

2

∆

sin

4

2 sin sin sin

ABC bh

ab C

s s a s b s c

rs

abc

R

R A B C

=

=

= − − −

=

=

=

� Common-sided triangles theorem: ∆

∆

ABC AM

A BC A M=

′ ′ (For M = AA’ ∩ BC)

� Angle bisector theorem: AB KB

AC KC= (For AK bisects ∠A and K on BC)

� Four centers of triangle:

� Centroid (G):

� Intersection of three medians.

� 2AG BG CG

A G B G C G= = =

′ ′ ′

� Circumcenter (O):

� Center of circumcircle.

� Intersection of three perpendicular bisectors of side.

� Orthocenter (H):

� Intersection of three altitudes.

� CDEH, CDAF, AFEH, BDAE, BDFH and CEBF are cyclic

quadrilaterals.

*

*

*

*

*

*

*

*

*

*


186

� H is orthocenter iff:

� AH ⊥ BC and

� (∠BAC + ∠BHC = 180°) or (BH ⊥ CA) or (CH ⊥ AB)

� Incenter (I):

� Center of incircle.

� Intersection of three internal angle bisectors.

� I is incenter iff:

� AI bisects ∠A and

� ( ) ( ) ( )12

or bisects or bisects BIC π BAC BI B CI C∠ = + ∠ ∠ ∠

� Two external angle bisectors intersect one internal angle bisector

at points called excenters.

� More about Cyclic Quadrilaterals

� Intersecting chords theorem:

(X = PQ ∩ RS) ⇒ (PQRS is cyclic ⇔ PX · QX = RX · SX)

� Ptolemy’s theorem: For ABCD a quadrilateral,

AB · CD + DA · BC ≥ AC · BD

Equality holds iff ABCD is cyclic.

� Geometric Transformation

� (Please also see coordinates geometric transformation in “Coordinates

Geometry”.) (Here point X, point Y, figure F and line L are mapped to X’, Y’,

F’ and L’; O is the invariant point.)

� Property of translation:

� F ≅ F’.

� L’ // L and L’ = L.

� XYY’X’ is parallelogram.

� Scaling:

� F ~ F’.

� L’ // L and L’ = |r| × L.

� O, X, X’ are collinear.

� Reflection:

� F ≅ F’.

� L’ = L and line of reflection bisects angle between L and L’.

*

*

*

*

*

*

*

*

*

*

*


187

� Rotation:

� F ≅ F’.

� L = L’ and angle between L and L’ is angle of rotation.

� ∆OXX’ ~ ∆OYY’

� ∆OXY ≅ ∆OX’Y’

� Spiral similarity (combination of rotation and scaling):

� F ~ F’.

� The angle between XY and X’Y’ is angle of rotation.

� ∆OXX’ ~ ∆OYY’.

� ∆OXY ~ ∆OX’Y’.

� Simple Solid Geometry

� Volume, surface area and total length of edges of objects:

Solid Volume Surface area Total length of edges

CUBE a3 6a

2 12a

CUBOID abc 2(ab + bc + ca) 4(a + b + c)

PRISM Ah − −

CYLINDER πr2h

2πr(r + h)

(Curved only: 2πrh) 4πr

PYRAMID 13Ah − −

CONE 21

3πr h

πr(r + s)

(Curved only: πrs)

( 2 2s r h= + )

2πr

SPHERE 34

3πr 4πr

2 0

ELLIPSOID 43πabc − 0


188

Self Assessment Exercises (The figures are not necessary drawn to scale.)

Level 1

1) Find the trigonometric functions of 60°.

2) If 7

cos15

A = and A ∈ (-90°, 0°), find 2

cot csc

sin tan

A A

A A

−.

3) Do the followings: Find the area of the shaded region below:

a) (Segment formula) Prove that the area of the segment shaded is

( )212

sinr θ θ− .

b) Hence, or otherwise, find the area of the shaded region below:

4) Let ∆ABC be a triangle and D, E, F on sides BC, CA and AB respectively such that

AD, BE, CF intersect at P. Prove that:

a) 1PD PE PF

AD BE CF+ + =

b) (Ceva’s theorem) 1AF BD CE

FB DC EA⋅ ⋅ =

5) Prove that 2

∆3 3

sABC ≤ and determine when will equality occur.

(s is the semi-perimeter.)

6) Do the followings:

a) (張角公式) Let ∆ABC be a triangle and P be a point. Prove that P is on BC if

and only if:

sin sin sinBAC BAP CAP

AP AC AB

∠ ∠ ∠= +

b) (MUMS-UMO 2000 Q18) In a right-angled triangle ∆ABD, AB = 4 and

BD = 3. A point C lies on BD such that 2 ∠BCA = 3 ∠BAD. Find the length

of BC.

1

1

θ r

r

*


189

7) Platonic solids are 3D object that every face is the same and is a regular polygon.

Cube and regular tetrahedron are examples of platonic solids. Briefly explain why

there are only five Platonic solids and describe their shape.

8) A sphere is inscribed inside a cylinder. Find:

a) Surface area of cylinder

Surface area of sphere

b) Volume of cylinder

Volume of sphere

Level 2

9) (PCMSIMC 2004/2 Q11) Let ∆ABC be a right-angled triangle with AB = 3,

BC = 4 and CA = 5. Perpendiculars are drawing to AC and BC repeatedly starting

from B. Find the area of shaded region below:

5

4

3

C B

A

10) (MUMS UMO 2001) Joe loved cantaloupes. He put a spherical one in his cubical

cantaloupe fridge and it just fit − touching all of the fridge’s (very thin) sides and

lid. However, an alien civilization angry at human worship of cantaloupes

destroyed all cantaloupe plantations on Earth. But Joe was prepared for an

escalation of anti-cantaloupist violence and had purchased a cantaloupe-fridge

cryogenic suspension device. To preserve his cantaloupe for future generations,

Joe put the fridge inside the spherical suspension chamber and it just fit. The

chamber wasn’t very well packed, so Joe decided to remove the fridge, puree all

his cantaloupes (all identical) and pour the puree directly into the chamber. How

many pureed cantaloupes can Joe fit in the cantaloupe suspension chamber? (The

answer need not be a whole number, and should be expressed exactly).

11) (HKMHASC 2004) Given D is a point on the diameter AB, where AD = 8, DB = 4

and CD ⊥ AB. Find the length of CD.

A B D

C


190

12) (PCMSIMC 2002/S Q6) In the figure, AB is a diameter of the circle and C is a

point on AB produced. CD is tangent to the circle at D and E is the foot of the

perpendicular from D to AB. If AD = 6 and ∠CDB = 30°, find the area of ∆DEB.

C E A B

D

13) Prove the converse of Ceva’s theorem. (See also Q4b)

[Hint: Let CP intersect AB at F’]

14) (HKCEE 2001/A Q15 Modified) Do the followings:

a)

The figure above shows a pyramid OPQR. The sides OP, OQ and OR are

mutually perpendicular to each other. Let S1, S2, S3 and S be the areas of

∆OPQ, ∆OQR, ∆ORP and ∆PQR respectively. Prove that

S2 = S1

2 + S2

2 + S3

2

b)

The figure above shows a cuboid ABCDEFGH. The lengths of AB, BC and

AF are 4, 3 and 2 respectively. A pyramid ABCG is cut from the cuboid along

the plane GAC. Find the distance from B to the plane GAC.

15) (TRML 2004/G Q1) Let H be the orthocenter of an acute-angled triangle ∆ABC. If

AH = 2 BC, find cos ∠A.

16) (AIME 1999 Q11) If 35

1

cot sin 5k

q k=

° = °∑ , find q.

P

Q R

O

2

4

3

A B

C D

E

F G

H

*

*


191

17) (KöMaL C. 524) Let ∆ABC be an isosceles triangle with AC = BC. P is a point on

AB such that ∠ACP = 30°. Q is a point outside the triangle which

∠CPQ = ∠CPA + ∠APQ = 78°. Given that all angles are integer measured in

degrees, find the angles of the triangles ∆ABC and ∆BPQ.

[Hint: ∠ACB > ∠ACP.]

18) The following is the net (“spread-out” form) of a solid.

a) Find tan 36° in surd form.

b) Find the total surface area of the solid in terms of x.

c) (HKMHASC 2004 Q6) After folded, which vertex will coincide with Z?

d) A decagon ABCDEFGHIJ is formed after folded. Find the value of

Surface area of solid

Area of decagon.

19) (IMOPHK 2003 Q13) Let ABCD be a square of side length 5. Let E be a point on

BC such that BE = 3 and CE = 2. Let P be a point on BD. Find the length of PB so

that PE + PC is minimum.

x

A

B

Z C D E

F

G

H

I

J


192

Answers

1) 3 1 3 2 3

sin 60 , cos60 , tan 60 3, cot 60 , sec60 2, csc602 2 3 3

° = ° = ° = ° = ° = ° =

2) –735⁄3872

3) b) 1 33

π− +

5) Equality occurs when ∆ABC is equilateral.

8) a) 3⁄2. b)

3⁄2

9) 96⁄41

10) 3 3

12) b) 12

61

15) 2

5

16) 5⁄2

17) ∠BPQ = 178°, ∠QBP = ∠PQB = 1°; ∠ABC = ∠CAB = 74°, ∠BCA = 32°.

18) a) 5 2 5− . b) 25 25 10 5x + . c) D. d) 2 5 .

19) 883⁄333


193

Chinese Translation of Important Terms Altitude 垂線

Amplitude 振幅

Angle at the center 圓心角

Angle at the circumference 圓周角

Angle bisector 角平分線

Angle bisector theorem 等分角定理

Angle in the alternate segment 交錯弓形之圓周角

Angles in the same segment 同弓形之圓周角

Arc 弧

Barycenter 重心

Center (of circle) 圓心

Center of mass 質心

Centroid 重心

Chord 弦

Circle 圓

Circumcenter 外心

Circumcircle 外接圓

Circumference 圓周

Circumradius 外接圓半徑

Circumscribed circle 外接圓

Common-sided triangles 共邊三角形

Common-sided triangles theorem 共邊比定理

Compound angle formulae 複角公式

Concave quadrilateral 凹四邊形

Concentric circles 同心圓

Concyclic 共圓

Cone 圓錐體

Congruence 全等

Converse 逆命題

Convex quadrilateral 凸四邊形

Cosecant 餘割

Cosine 餘弦

Cosine rule 餘弦定理

Cotangent 餘切

Cube 立方體


194

Cuboid 長方體

Curved surface 曲面

Cyclic quadrilateral 圓內接四邊形

Cylinder 圓柱體

Degenerated 退化的

Diagonal 對角線

Diameter 直徑

Disc 圓盤

Double angle formulae 倍角公式

Edge 邊

Ellipsoid 橢球

Equal circles 同圓

Equal ratios theorem 等比定理

Equilateral triangle 等邊三角形

Equilibrium position 等位

Excenter 旁心

Excircle 旁切圓

Exradius 旁切圓半徑

Exterior angle 外角

Face 面

Foot of perpendicular 垂足

Geometry 幾何學

Half angle formulae 半角公式

Heron’s formula 希羅公式

Hexahedron 六面體

Homothety 位似變換

Incenter 內心

Incircle 內切圓

Inradius 內切圓半徑

Inscribed circle 內切圓

Intercept 截距

Intercept theorem 截距定理

Interior angle 內角

Intersecting chords theorem 交弦定理

Isosceles trapezoid 等腰梯形

Isosceles triangle 等腰三角形

Lateral face 側面


195

Major ~ 優 ~

Median 中線

Midpoint theorem 中點定理

Minor ~ 劣 ~

Necessary condition 必要條件

Nets 展開圖

Orthocenter 垂心

Parallelogram 平行四邊形

Period 周期

Perpendicular bisector 垂直平分線

Phase 相角

Planar geometry 平面幾何

Platonic solid 柏拉圖多面體

Polar 極線

Pole 極點

Polygon 多邊形

Prism 柱體

Product-to-sum formulae 積化和差公式

Ptolemy’s theorem 托勒密定理

Pyramid 錐體

Pythagoras’ theorem 畢氏定理

Quadrilateral 四邊形

Radian 弧度

Radius (pl. radii) 半徑

Rectangle 長方形

Rhombus 菱形

Right angle 直角

Right pyramid 直立錐體

Right-angled triangle 直角三角形

Secant 正割

Sector 扇形

Segment 弓形

Semi-axis 半軸

Semicircle 半圓

Semi-perimeter 半周長

Similarity 相似

Sine 正弦


196

Sine rule 正弦定理

Skew quadrilateral 扭曲四邊形

Slant height 斜高

Solid geometry 立體幾何

Solving the triangle 解三角形

Sphere 球體

Spheroid 扁球體

Spiral similarity 旋轉位似變換

Square 正方形

Subsidiary angle form 輔角式

Sufficient condition 充分條件

Sum-to-product formulae 和差化積公式

Supplementary 互補

Surface area 表面積

Symmetric 對稱

Tangent 正切

Tangent property 切線性質

Tangent-chord angle 弦切角

Tangential quadrilateral 圓外切四邊形

Trapezium (Br.) 梯形

Trapezoid (Am.) 梯形

Triangle inequality 三角不等式

Trigonometric function 三角函數

Trigonometry 三角學

Volume 體積


197

References:

� Wikipedia, the free encyclopedia. [http://en.wikipedia.org/]

� MathWorld – A Wolfram Web Resource. [http://mathworld.wolfram.com/]

� Mathematics Database. [http://www.mathdb.org/]

Original documents:

� Introduction to Trigonometry: Chapter 8: “Trigonometry” Section 1, 2, 3, 4, 5.

� Trigonometry in Depth: Chapter 8: “Trigonometry” Section 8, 9, 10, 11.

� Lines and Line Segments: Chapter 0: “Before you start”; Chapter 9:

“Geometry” Section 2, 4.

� Triangles: Chapter 9: “Geometry” Section 1, 3.

� Circles: Chapter 9: “Geometry” Section 7.

� Quadrilaterals: Chapter 9: “Geometry” Section 5, 6, 8.

� More about Triangles: Chapter 9: “Geometry” Section 1.

� More about Cyclic Quadrilaterals: Chapter 9: “Geometry” Section 8, 9.

� Geometric Transformation: Chapter 17: “More about Geometry”.

� Simple Solid Geometry: Chapter 18: “Solid Geometry”.

Documents

Mathematics Competition Training Class Notes Elementary ...ylmass.edu.hk/~mathsclub/MCTCN/ege.pdf · Mathematics Competition Training Class Notes Elementary Geometry 119 Elementary