Download Geometry

Advanced Mathematics Training Class Notes
Chapter 9: Geometry
A
Chapter 9
Geometry
c
Triangle
b
B
a
C
We often denote the opposite side of A as “a”, B as “b” and C as “c”.
If we want to say the area of ∆ABC, we will write it as ∆ABC or [ABC].
Triangle is a polygon with three sides. There are three types of triangles regarding the
(largest) angle:
The four centers of a triangle
A
Acute-angled triangle
(銳角三角形)
Right-angled triangle
(直角三角形)
hA
Obtuse-angled triangle
(鈍角三角形)
All three angles of an acute-angled triangle is acute, i.e., less than 90°.
One angle of a right-angled triangle is a right angle, i.e., 90°.
One angle of an obtuse-angled triangle is obtuse, i.e., greater than 90°.
HA
A line passing through a vertex and perpendicular to the side opposites to it is called an
“altitude” (垂線), also know as “height”. In the figure, AHA is an altitude. In a triangle, there
are three altitudes altogether and they are concurrent. The point they intersect is called the
“orthocenter” (垂心), often denoted as “H”.
A
Beside, there are also two kinds of triangles considering of the sides:
mA
Isosceles triangle
(等腰三角形)
Equilateral triangle
(等邊三角形)
MA
A line passing through a vertex and cuts the corresponding side in half is called a “median”
(中線). In the figure, AMA is a median. In a triangle, there are three medians and they
intersect at a point called “centroid” (重心), written as “G”, which means “center of mass”.
Two sides of an isosceles triangle are equal.
All sides of an equilateral triangle are equal.
If a triangle has vertices A, B and C, we call this triangle “∆ABC”.
59
Advanced Mathematics Training Class Notes
Chapter 9: Geometry
Angles of a triangle
A
The sum of all interior angles (內角) of a
triangle, i.e., ∠CAB , ∠ABC and ∠BCA , is
always equals to 180° (π radians).
∠A + ∠B + ∠C = 180°
[Reference: ∠ sum of △ ]
iA
IA
An angle bisector (角平分線) is a line that pass through a vertex and divides the angle there
in half. Three angle bisectors are there in a triangle, and they intersect as the incenter (內心),
shorted as “I”.
A
E
B
Moreover, an exterior angle (外角), ∠EBA , is
the sum of the two angles not next to it, i.e. ∠BCA and ∠CAB.
∠EBA = ∠A + ∠C
[Reference: ext. ∠ of △ ]
C
Pythagoras’ Theorem (畢氏定理
畢氏定理)
畢氏定理
For a right-angled triangle, the square of the
hypotenuse is equals to the sum of the square of
the two other sides.
c2 = a2 + b2
[Reference: Pyth. Thm.]
OA
A line that bisects a side and perpendicular to it is called a “perpendicular bisector” (垂直
平分線). As you expected, three perpendicular bisectors join to form the “circumcenter” (外
心), denoted as “O”.
In contrast, in a triangle, if the square of one side is
equals to the sum of square of two other sides, it
must be right-angled.
If c2 = a2 + b2, then ∆ABC is rt.∠ ∆.
[Reference: Converse of Pyth. Thm.]
Two circles of a triangle
All triangles are inscribed by a circle (the green one), and are inscribing a circle (the red one).
The red circle is called the “inscribed circle” (內接圓). Its center is the incenter.
The green circle is called the “circumscribed circle” (外接圓). Its center is the circumcenter.
60
c
b
a
Advanced Mathematics Training Class Notes
Chapter 9: Geometry
Trigonometry in general triangles
Congruent and similar triangles (全等及相似三角形
全等及相似三角形)
全等及相似三角形
We have come across how trigonometric functions deal with right-angled triangles. Now we
extend the use of them to general triangles.
Two figures are said to be congruent (全等) if they have the same shape and size.
A
In any triangle ∆ABC,
a
b
c
=
=
= 2R
sin A sin B sin C
[Reference: sine rule]
2
2
c
2
c = a + b − 2ab cos C
B
Two figures are said to be similar (相似) if they have the same shape and their sizes are in
ratio.
C
a
a2 + b2 − c2
cos C =
2ab
[Reference: cosine rule]
The above two figures are congruent. Note that congruency is not affected by rotation.
b
The above two figures are similar.
“R” is radius of the circumscribed circle.
A
Example: Solve the following triangle.
By cosine rule,
a 2 = 32 + 62 − 2 ( 3)( 6 ) cos 45°
45°
6
Two triangles are congruent if any one of the following conditions is met:
(In the references, “S” means “side”, “A” means “angle”)
All corresponding sides are the same [SSS]
Two corresponding sides are the same, and also their included angle (夾角) [SAS]
Two corresponding angles are the same, and also
The sides between them [ASA], or
The sides not between them [AAS]
Both triangles are right-angled. The hypotenuses are the same, and also another pair of
sides. [RHS]
3
a = 45 − 18 2
By sine rule,
B
3
6
45 − 18 2
=
=
sin B sin C
sin 45°
3
6
∴ sin B =
, sin C =
2 45 − 18 2
2 45 − 18 2
(
)
(
a
C
)
SSS
If you like, you can evaluate the results by a calculator (although you can’t), giving a = 4.42,
B = 106.32°, C = 28.68°. Note that another solution for B, 73.68°, is dropped because that
will make the angle sum < 180°. With similar reason, 151.32° for C is rejected as the angle
sum in this case will be > 180°.
If you don’t want to see doubt cases, use cosine rule.
SAS
ASA
AAS
Note that “ASS” is not a condition of congruent triangles.
RHS
61
Advanced Mathematics Training Class Notes
Chapter 9: Geometry
Two triangles are similar if any one of the following conditions are met:
All corresponding angle are the same [AAA]
All corresponding sides are proportional (成比例) [3 sides prop.]
Two corresponding sides are proportional, and their corresponding included angles are
the same. [2 sides ratio, inc. ∠]
AAA
3 sides prop
Isosceles and equilateral triangles
If ∆ABC is isosceles (AB = AC), it follows:
∠B = ∠C [base ∠s, isos. ∆]
The altitude, median, angle bisector and perpendicular
bisector corresponding to A represents to same line.
All these conditions are conversable. That is, if ∠B = ∠C, then
AB = AC.
[Reference: side. opp. eq. ∠s]
2 sides ratio,
inc. ∠
P
C
Q
A
c
However, this is not convenient in most cases. Since
h = b sin C , we get:
∆ABC = ab sin C / 2
R
If ∆ABC ≅ ∆PQR:
AB = PQ, BC = QR, CA = RP.
[Ref: corr. sides, ≅ ∆s]
∠A = ∠P, ∠B = ∠Q, ∠C = ∠R.
[Ref: corr. ∠s, ≅ ∆s]
B
h
b
a
Besides these two, there are also massive ways to obtain the area.
△ ABC = 12 ah
= 12 ab sin C
If ∆ABC ∼ ∆PQR,
AB BC CA
=
=
PQ QR RP
[Ref: corr. sides, ∼ ∆s]
∠A = ∠P, ∠B = ∠Q,
[Ref: corr. ∠s, ∼ ∆s]
C
Area of triangle
In primary mathematics, we learnt that:
Area = 1/2 (Base) (Altitude)
∆ABC = ah/2
B
B
If DABC is equilateral, then the theorems for isosceles triangles apply, in addition to:
∠A = ∠B = ∠C = 60°.
The orthocenter, centroid, incenter and circumcenter represent the same point.
If ∆ABC and ∆PQR are congruent, it is denoted as ∆ABC ≅ ∆PQR.
If ∆ABC and ∆PQR are similar, it is denoted as ∆ABC ∼ ∆PQR.
A
A
= s ( s − a )( s − b )( s − c )
This is called the “Heron’s Formula”
= rs
=
∠C = ∠R.
abc
4R
= 2 R 2 sin A sin B sin C
=
62
a 2 sin B sin C
2sin ( B + C )
(Note:
s = 12 ( a + b + c ) is the semi-perimeter.
r is the radius of the inscribed circle.
R is the radius of the circumscribed circle.)
C
Advanced Mathematics Training Class Notes
Chapter 9: Geometry
Common side triangles theorem (共邊定理
共邊定理)
共邊定理
Parallel Lines
If two triangles ∆ABP and ∆ABQ share a common side AB,
△ ABP MP
=
△ ABQ MQ
This is known as the “common side triangles theorem”, and is useful in tackling many
geometric problems.
P
P
P
Q
M
B
Q
C
P
B
A
M
Q
M
B
Example: In ∆ABC, X is a point on BC, Y is a point on CA and Z is a point AB, such that
BX:XC = CY:YA = AZ:ZB = 2:1. Lines AX, BY and CZ join to form the triangle ∆PQR. Find
A
∆PQR : ∆ABC.
Using the given conditions, first find
△ ABC △ ABR +△ RBC +△ ARC
=
△ RBC
△ RBC
AY
AZ
=
+1+
CY
BZ
= 12 + 1 + 2 = 72
△ RBC △ APC
,
△ ABC △ ABC
and
△ ABQ
△ ABC
F
Note that the statement “AB // CD” should change as the names of the parallel lines change.
The above three theorems are conversable (可逆), i.e., if EF cuts two lines AB and CD at P
and Q, AB // CD if:
The corresponding angles (對應角) are equal [corr. ∠s equal]
The alternate angles (錯角) are equal [alt. ∠s equal]
The sum of two interior angles is 180°. [int. ∠s supp.]
.
Q
Y
Z
R
D
If a line EF cuts the parallel lines AB and CD at P and Q respectively, it is true that
∠EPA = ∠EQC.
∠EPB = ∠EQD.
∠APF = ∠CQF.
∠FPB = ∠FQD. [corr. ∠s, AB // CD]
∠APQ = ∠PQP.
∠BPQ = ∠CQP. [alt. ∠s, AB // CD]
∠BPQ + ∠DQP = 180°.
∠APQ + ∠CQP = 180°. [int. ∠s, AB // CD]
Q
A
B
A
B
A
P
E
Q
M
A
Parallel lines are a pair of lines, which they do not intersect with each other. If AB is parallel
to CD, we write AB // CD. We put arrows “→” on the lines to show that they are parallel.
P
B
C
X
Therefore, ∆RBC : ∆ABC = 2 : 7. With the same reason, ∆APC:∆ABC = ∆ABQ:∆ABC = 2:7.
Thus, ∆PQR : ∆ABC = 1:7.
63
Advanced Mathematics Training Class Notes
Chapter 9: Geometry
Polygons
Intercepts
For general convex polygons (凸多邊形), if it has n sides (it is an n-gon), the sum of all its
interior angles is 180°(n – 2). For example, a pentagon (五邊形) has an interior angle sum of
180°(5 – 2) =540°.
The sum of all exterior angles of a polygon is always 360°.
A
P
Name
Sides
Quadrilateral 4
Pentagon
5
Hexagon
6
Heptagon
7
Octagon
8
Nonagon
9
Decagon
10
An n-gon always has
1n
2
B
[Reference: mid-point theorem]
In ∆ABC, if P is the mid-point of AB and PQ // BC such that Q is on AC, then:
AQ = QC
[Reference: intercept theorem]
For regular polygons (正多邊形), i.e., Polygon with all sides and angles the same, if there
are n sides, and length of each side is l,
180° ( n − 2 )
Size of each interior angle =
n
360°
Size of each exterior angle =
n
nl
 180° 
Area of the polygon =
cot 

4
 n 
More generally, if R and S lie on AB and AC respectively:
AR AS
If RS // BC, then
=
[equal ratios theorem]
RB SC
AR AS
If
=
, then RS // BC [converse of equal ratios theorem]
RB SC
If AP // BQ // CR // DS, and AB = BC = CD, it follows
PQ = QR = RS.
[Reference: equal intercepts]
l
 180° 
cot 

2  n 
Radius of inscribed circle =
Radius of circumscribed circle =
C
In ∆ABC, if P and Q are mid-points of AB and AC respectively, then:
PQ // BC,
PQ = 12 BC
( n − 3) diagonals (對角線)
Q
A
l
 180° 
csc 

2
 n 
B
C
D
64
P
Q
R
S
Advanced Mathematics Training Class Notes
Chapter 9: Geometry
Quadrilaterals (四邊形
四邊形)
四邊形
Parallelograms
Quadrilateral is a polygon with 4 sides. It can be divided into many subtypes.
Cyclic quadrilateral (圓內接四邊形) is a class of quadrilaterals that having a circumscribed
circle.
Tangential quadrilateral ( 圓外切四邊形 ) is a class of quadrilaterals that having an
inscribed circle.
Trapezoid (梯形) is a quadrilateral that one pair of opposite sides is parallel.
Parallelogram (平行四邊形) is a quadrilateral that both pairs of opposite sides are parallel.
Parallelogram is quadrilateral that both pairs of opposite sides are parallel. The word
“parallelogram” is often shortened as “// gram”.
For any convex quadrilaterals, the interior angle sum is 360°.
[Reference: ∠ sum of quad.]
If the two diagonal is drawn, they divides the quadrilateral into
four parts. Here they are called α, β, γ and δ respectively.
Referring to the figure, there is:
αγ = βδ
If it is a trapezoid, and the parallel sides are parts of
region β and δ, then:
α = γ = βδ
A parallelogram is a subtype of trapezoid.
Consider the parallelogram ABCD, as shown. The properties of parallelogram tell us that:
AB // DC
A
AD // BC
AB = DC
AD = BC [opp. sides, // gram]
∠B = ∠D
E
∠A = ∠C [opp. ∠s, // gram]
AE = EC
DE = EB [diags, // gram]
∆ABC = ∆ADC
D
C
∆BAD = ∆BCD [diag bisects area]
δ
α
γ
β
B
To prove a quadrilateral ABCD is a parallelogram, we need to show any of the followings:
AB // DC and AD // BC
AB = DC and AD = BC [opp. sides equal]
AB // DC and AB = DC [2 sides equal and //]
AD // BC and AD = BC [2 sides equal and //]
∠B = ∠D and ∠A = ∠C [opp. ∠s equal]
AE = EC and DE = EB [diags. bisect each other]
If a quadrilateral is tangential, the sum of both the opposition sides are equal.
a+c=b+d
This theorem is conversable.
The study of parallelogram and cyclic quadrilateral will start later in this chapter.
Rectangle (長方形
長方形)
長方形
Rectangle is a special type of parallelogram that all of its angles are right angles. Besides the
properties of parallelogram, rectangle also exhibits the following properties:
∠A = ∠B = ∠C = ∠D = 90°
AC = BC
AE = BE = CE = DE
It is a cyclic quadrilateral
[Reference: property of rectangle]
65
Advanced Mathematics Training Class Notes
Chapter 9: Geometry
Rhombus (菱形
菱形)
菱形
semi-circle (半圓) is an arc equal to half the circumference.
Rhombus is a kind of parallelogram that all sides of it are equal. Rhombus shows the
property of parallelogram as well as the followings:
AB = BC = CD = DA
AC ⊥ BD
∠ABD = ∠CBD = ∠ADB = ∠CDB
∠CAB = ∠CAD = ∠ACB = ∠ACD
It is a tangential quadrilateral
[Reference: property of rhombus]
, which means the minor arc from point B to C. If not specified, The red curve is BC
XY
usually denote the minor arc. If major arc is need, write BAC .
A sector is a region that is enclosed by two radii and an arc, like the yellow part. A segment
(弓形) is a region enclosed by a chord and an arc, like the red part.
The sector and segment can also be divided into major and minor according to their area.
All circles are similar.
Square (正方形
正方形)
正方形
Two circles are congruent if they have the same radii. They are called “equal circles”.
Square, or regular quadrilateral, is a combination of rectangle and rhombus. It has both the
properties of rectangle and rhombus.
Circles sharing common center but different radii are called “concentric circles” (同心圓).
An inscribed circle (內切圓) of a polygon is a circle that touches every
sides of the polygon at one point only.
Circle
A circle is a collection of points that is equidistance (同距) from a point called “center” (圓
心). We name the circle according to their center, e.g., the circle O:
B
A circumscribed circle (外接圓) of a polygon is a circle that every
vertices of the polygon lie on this circle.
A
C
O
Chords
If a straight line is drawn passing through the center and the
mid-point of a chord, it can be seen that the line is also
perpendicular to that chord. In fact:
If ON ⊥ AB, then AN = NB.
[line from center ⊥ chord bisects chord]
If AN = NB, then ON ⊥ AB.
[line joining center to mid-pt of chord ⊥ chord]
The circular curve is called the circumference (圓周). Any line connecting the center and a
point on the circumference (e.g. OA) is called the radius (半徑) (pl. radii). Any line
connecting two points on the circumference (e.g. BC) is called the chord (弦). The chord
passing through the radius is the diameter (直徑). An arc (弧) (e.g. The red curve) is a part
of the circumference.
An arc is further classified into 3 types. A minor arc (劣弧) is an arc shorter than half the
circumference. A major arc ( 優弧 ) is an arc longer than half the circumference. A
66
O
A
N
B
Advanced Mathematics Training Class Notes
Chapter 9: Geometry
Moreover, two chords are equal if and only if they are
equidistance from the center.
D
H
If AB = CD,
A
O
A
OM = ON
[Reference: equal chords, equidistance from center]
Relationships between arc, chord, angle, segment and sector
C
M
N
B
If OM = ON,
O
B
From Chapter 8, we knew that if ∠AOB is measured in radian, then
AB = rθ
AB = CD
[Reference: chords equidistance from center equal]
Angles
AB
Area of sector AOB = 12 r 2θ = 12 r ⋅ P
An angle subtended by an arc or a chord at the center is
called the angle at center (圓心角), e.g. ∠AOB. Similarly, if
the angle is subtended at the circumference, it is called the
angle at circumference (圓周角), e.g. ∠APB.
With r is the radius = OB, θ = ∠AOB.
It can also be seen that the area of segment ABH is
θ
1 r2
2
O
2θ
The angle at center is twice the angle at circumference if is
subtended by the same arc / chord.
∠AOB = 2 ∠APB
Example: In the above figure, ∠AOB = 120°, OB = 5 cm, find the area of segment ABH.
120° = 2π/3. Apply the segment area formula, the area of segment ABH is:
1  2π
2π
( 5)  − sin 
2  3
3 


5 2π
3
= 
−

2 3
2 
A
B
[Reference: ∠ at center twice ∠ at ⊙ ce ]
This theorem also applies to these two cases:
If AB is the diameter, ∠APB is then a right angle.
[Reference: ∠ is semi-circle]
By angle at center twice angle at circumference, we got
angles inside the same segment are equal.
(θ − sin θ )
This is known as the segment area formula.
(
5 4π − 3 3
)
cm 2
12
Which is approximately 3.071 cm2.
=
Q
P
∠APB = ∠AQB
[Reference: ∠s in same segment]
B
A
67
Advanced Mathematics Training Class Notes
Chapter 9: Geometry
In a circle, the arc, chord, angle (no matter it is at center or circumference), sector and section
are closely related. If there are two sets of them in the same circle, and any one of them are
the same, then also the others.
B
For example, if the two angles at center are the same, then also:
The two chords
The two arcs
The two segments (green regions)
The two sectors (green and yellow regions)
All of them are conversable.
C
P
Q
If PQ is tangent to circle passing ABC,
∠QAC = ∠ABC
∠PAB = ∠ACB
[Reference: ∠ in alt. segment]
This theorem is conversable.
If two angles are in ratio r:s, then also
The two arcs
The two sectors (green and yellow regions)
Again, they are conversable.
X
Tangent
A tangent (切線) of a circle is a line that cuts the circle at one point
only. In this figure, XU and XT are two tangents of the circle O.
U
The points T and U are called the point of contact (切點). A
tangent is always perpendicular to the radius, i.e.,
If XT is tangent to circle O, and T is the point of
O
contact, then
XT ⊥ TO
[Reference: tangent ⊥ radius]
This theorem is conversable.
A
Cyclic Quadrilateral
A cyclic quadrilateral is a quadrilateral that has a circumscribed circle.
The figure on the right shows a cyclic quadrilateral. For any cyclic quadrilateral ABCD, it is
A
true that:
∠A + ∠C = ∠B + ∠D = 180° [opp. ∠s, cyclic quad.]
∠ABC = ∠ADE [ext. ∠s, cyclic quad.]
E
B
T
To prove a quadrilateral is cyclic, any one of the
following conditions must be met:
∠ACD = ∠DBA or etc.
[converse of ∠s in same segment]
∠A + ∠C = 180° or
∠D + ∠B = 180° [opp. ∠s supp.]
∠ABC = ∠ADE [ext. ∠ = int. opp. ∠]
If two tangents are drawn from the point X to the circle O, with points of
contact T and U, then:
XT = XU
∠XOT = ∠XOU
∠OXT = ∠OXU
[Reference: tangent properties]
D
The area of a cyclic quadrilateral with sides a, b, c and d =
The point X is called the pole (極點), and the line UT is called the polar (極線).
where s is the semi-perimeter =
68
1
2
(a + b + c + d ) .
C
( s − a )( s − b )( s − c )( s − d ) ,
Advanced Mathematics Training Class Notes
Chapter 9: Geometry
Euler’s Inequality
Intersecting Chords Theorem (交弦定理
交弦定理)
交弦定理
Let R be the radius of the circumscribed circle of a triangle, and r be the radius of the
inscribed circle. It follows:
A
R > 2r.
Equality holds if and only if the triangle is equilateral
D
A’
A’
A
B’
X
Ptolemy’s Theorem
X
B
For general point A, B, C and D:
AB · CD + AD · BC > AC · BD
Equality holds if and only if ABCD is a cyclic quadrilateral.
(This can be regarded as a condition for cyclic quadrilateral.)
A
B’
B
If two chords AB and A’B’ of circle O intersects at a point X, then
XA · XB = XA’ · XB’
This is the intersecting chords theorem.
2
2
B
Bearings (方位
方位)
方位
This theorem in conversable. If AB and A’B’ are two lines intersecting at point X, and also
XA · XB = XA’ · XB’, then ABA’B’ is a cyclic quadrilateral.
2
C
Bearing is a way to tell the direction of an object from a given point on the same horizontal
plane. There are mainly two kinds of bearings – compass bearing (羅盤方位角) and true
bearing (真方位角).
2
Moreover, XA · XB = | OX – r |, where r is the radius of the circle. The value OX – r is the
power of point X with respect to circle O.
Compass bearing
In addition, if A’ and B’ represents the same point, T, i.e., XT is tangent to O, then
XA · XB = XT2
[Reference: tangent property]
In using the compass bearing, directions are measured either from North or South.
A
N
Geometric Inequalities
B
45°
Geometric inequalities are inequalities with geometric conditions added.
60°
Triangle Inequality
W
E
O
If a, b and c are sides of a triangle, then
a+b>c
b+c>a
c+a>b
25°
C
69
S
Advanced Mathematics Training Class Notes
Chapter 9: Geometry
In the figure on the last page, if a person stands on O facing North (N), then turn 45° towards
East (E), he will see the point A. We call the bearing from O to A is N45°E. Similarly, the
bearing of O to B is N60°W, and the bearing from O to C is S25°W.
Techniques in Doing Geometric Problems
Identifying similar / congruent / common side triangle
True Bearing
In figures involving many triangles, try to identify as many similar / congruent / common
side triangles as you can. Then apply the properties of similar / congruent triangles, or
common side triangles theorem. This helps finding the answer a lot.
In true bearing, directions are measure from the North only in clockwise direction.
A
N
B
45°
300°
W
Example: (PCMSIMC 2002) In the figure, ABCD is a parallelogram. E is a point on AD, F is
a point on AB and G is a point on BD such that DEFG is a rectangle. If BG = 1, CD = 8 and
FD ⊥ CD, find the length of AE.
B
C
E
O
205°
F
G
C
Again referring to this figure, assume there is a person at O facing North. If he turns 205°
clockwise, he can see point C. We say that the bearing from O to C is 205°. Likewise, the
bearing from O to B is 300°, and from O to A is 045°. Note that when expressing true bearing,
there are always 3 digits.
A
When referring to “bearing”, it usually means compass bearing.
70
E
D
Advanced Mathematics Training Class Notes
Chapter 9: Geometry
B
F
That is,
C
DE
=
AE
DE
1
AE
DE
G
AE = DE 3
AE
So,
8
(1 + DE ) + ( DE
2 2
3
+ DE
)
2
= 82
DE 6 + 3DE 4 + 3DE 2 + 1 = 64
( DE + 1)
3
2
A
E
D
= 64
2
DE + 1 = 4
DE = 3
Add the suitable information first.
Similar triangles are all right-angled triangles you can see.
Therefore,
AE =
So many similar triangles! Let’s do the search with this clue.
Consider ∆BFG and ∆FAE.
AE FE
=
(corr. sides, ~∆s)
FG
1
That means
AE
= GD (opp. sides, // gram)
DE
Also,
BC = AE + ED (opp. sides, // gram)
Thus,
( 3)
3
=3 3
If you didn’t consider the similar triangles, you may not able to get AE = DE3 and get stuck.
Adding additional lines (輔助線
輔助線)
輔助線
Additional lines can help you figure out some “hidden” clues. Example presented in page 5.
Removing unused lines
Sometimes the lines are there to block your vision, or not useful at that time. Simply remove
them to clear your way.
2
AE 

2
2
1 +
 + ( AE + DE ) = 8 (Pyth. Thm.)
DE


No good in expanding it. Instead, Consider ∆FED and ∆AEF,
DE FE
=
FE AE
Use coordinates
Sometimes it is much more easier when a geometric figure is put on a suitable coordinate
plane. Consider the following example:
A
E
B
X
F
D
71
C
Advanced Mathematics Training Class Notes
Chapter 9: Geometry
ABCD is of course a square, and BF:FC = 2:1. Suppose you want to find how the intersecting
point of BD and AE (X) divides BD.
Now we “construct” a point Q such that BCPQ is a parallelogram.
P
How to do? First, set up a coordinate system, with the origin at D. So D is (0, 0). Define the
length of the side of the square ABCD be 6 (Do you know why?). So A is (0, 6), B is (6, 6), C
is (6, 0), E is (3, 6) and F is (6, 2). Consider ∆BEF and ∆DEF.
D
C
Q
6 6
1 3 6
△ BEF =
=6
2 6 2
6 6
△ DEF =
0 0
3 6
6 2
0 0
A
B
= 15
It seems to be messier. But observe that:
∠QAB = ∠PDC
( corr. ∠s, ≅△)
( given )
( alt. ∠s, BC // PQ )
= ∠PBC
BX △ BEF
6 2
=
=
=
XD △ DEF 15 5
Therefore, X divides BD in ratio 2:5. A piece of cake! Isn’t it? Imagine how you can find the
area of ∆DEF without help of coordinate system!
By common side triangles theorem,
= ∠QPB
So ABQP is a cyclic quadrilateral (converse of ∠s in alt. segment)
Thus:
∠APB = ∠AQB
( ∠s in alt. segment )
Constructing something from nothing (無中生有
無中生有)
無中生有
= ∠DPC
Like adding additional lines, but this time you create the clue yourself.
Example: (GR) If lines PB and PD, outside a parallelogram ABCD, make equal angles with
the sides BC and DC (i.e., ∠PBC = ∠PDC), respectively, as in the figure, then prove that
∠CPB = ∠DPA. (Don’t be confused. This is not a 3D figure!)
( corr. ∠s,
≅△ )
∠CPB = ∠APB − ∠APC
And finally, ∠DPA = ∠DPC − ∠APC
∴∠CPB = ∠DPA
( Q.E.D.)
Constructing something from nothing is helpful, but it is difficult to practice.
P
Measure it!
D
A
C
It you can’t really get rid of the problem, draw it out and measure it! It can also used to
ensure if your answer is correct or not.
Practice makes the best
B
And yes, practice more can help to memorizing some specific problems, and work them out
fast. This also applies to other types of competition, or even in your life.
72
Advanced Mathematics Training Class Notes
Chapter 9: Geometry
2. (HKCEE 1992) In the figure, A, B, C, D, E and F are points on a circle such that AD // FE
=
AFE . AD intersects BE at X. AF and DE are produced to meet at Y.
and BCD
Revision
a)
b)
c)
d)
In this long chapter, we’ve learnt:
1. Theories about a triangle
2. Similar and congruent figures
3. Theories about parallel lines
4. Theories about polygons
5. Theories about intercepts
6. Theories about general quadrilaterals
7. Theories about parallelograms
8. Theories about circles
9. Theories about cyclic quadrilaterals
10. Intersecting chords theorem
11. Some geometric inequalities
12. Techniques in doing geometric problems
Prove that ∆EFY is isosceles.
Prove that AB // DE.
Prove that AXEY is a cyclic quadrilateral.
If ∠ABE = 47°, find ∠EFY, ∠AXE and ∠DYA.
A
B
F
C
X
Y
E
D
3. (HKMO 2001 Heat) In the figure, ABEF and BCDE are squares, BE = 6 cm and
are arcs drawn with centers F and D respectively. Find the total area of the
AE and CE
shaded parts.
Exercise
A
B
C
In the followings, if not specified, x is the variable. If ∆ABC is a triangle, its sides are a, b, c
corresponding to vertices A, B and C. Its radius of inscribed circle is denoted as r. Its radius
of circumscribed circle in denoted as R.
1. If ∆ABC is a triangle, prove that:
a) sin C = sin (A + B)
sin A + 2sin B + 3sin C
1
b)
=
a + 2b + 3c
2R
c) sin 2 A + sin 2 B − sin 2 C = 2sin A sin B cos C
d) △ ABC = rs
abc
e) (HKCEE 2000 Partial) R =
4a 2b 2 − a 2 + b 2 − c 2
(
)
F
E
D
4. (IMO Prelim HK 2003) Given a rectangle ABCD, X and Y are respectively points on AB
and BC. Suppose the areas of triangles ∆AXD, ∆BXY and ∆DYC are respectively 5, 4 and
3. Find the area of ∆DXY.
5. (IMO Prelim HK 2003) A man chooses two positive integers m and n. He then defines a
positive integer k to be good if a triangle with side lengths log m, log n and log k exists.
He finds that there are exactly 100 good numbers. Find the maximum possible value of
mn.
6. (IMO Prelim HK 2003) Let ∆ABC be an acute-angled triangle, BC = 5. E is a point on AC
such that BE ⊥ AC, F is a point on AB such that AF = BF. Moreover, BE = CF = 4. Find
the area of the triangle.
7. (IMO Prelim HK 2003) The ratio of the sides of a triangle, which is inscribed in a circle
of radius 2 3 , is 3 : 5 : 7. Find the area of the triangle.
2
73
Advanced Mathematics Training Class Notes
Chapter 9: Geometry
12. In ∆ABC, F is a point on AB, D is on BC and E is on CA. If AD, BE and CF meet at a
point P, prove that:
PD PE PF
+
+
=1
a)
AD BE CF
AF BD CE
b)
⋅
⋅
= 1 [Ceva’s Theorem]
FB DC EA
13. Let a, b, c be positive real numbers such that a2 + b2 – ab = c2. Prove that (a–c)(b–c) < 0.
14. (IMO 1988) ∆ABC is a triangle right-angled at A, and D is the foot of the altitude from A.
The straight line joining the incenters of ∆ABD, ∆ACD intersects the sides AB, AC at
points K and L respectively. Show that ∆ABC > 2 ∆AKL.
BD AB
15. Show that, in ∆ABC, if the angle bisector of ∠A intersects BC at D, then
=
.
CD AC
[Angle Bisector Theorem]
16. (ISMC 2000 Final) Given 17, 25 and 26 are the three sides of a triangle and θ is the angle
θ a
opposite to the side 17. Given tan = , where (a, b) = 1. Find a + b.
2 b
8. (PCMSIMC 2003) In the figure, ABCD is a rhombus. X and Z are points on AD and BC
respectively. O is a point inside the rhombus, Y is a foot of the perpendicular from O to
XZ and the distance from O to AB, BC, CD and DA are all equal to OY. If ZX and BA are
produced to meet at W, WX = 5, XY = 3 and YZ = 2, find the area of ABCD.
[Image is on the next page]
W
A
B
X
O
Y
D
Z
C
9. (PCMSIMC 2003) In ∆ABC, AB = 15, BC = 14 and CA = 13. X and Y are the feet of the
perpendiculars from A to BC and C to AB respectively. Find the length of XY.
10. (PCMSIMC 2002) In the figure, ABX and ACY are straight lines, XC touches the circle at
C, XC = XY and AX ⊥ XY. If AB = 2, BX = 1, find the length of AY.
X
B
Y
11. Show that △ ABC ≤
C
A
s2
. Determine when equality holds.
3 3
74
Advanced Mathematics Training Class Notes
Chapter 9: Geometry
Suggested Solutions for the Exercise
2d) 47°, 86°, 94°.
3) 18π – 18
4) 2 21
5) 134
6) 8 3 − 6
135
7)
3
49
8) 80
9) 39/5
10) 2 3
11) When the triangle is equilateral
16) 23
75