F.3 Mathematics Supplementary Notes
Chapter 5 More on Deductive Geometry
Chapter 5 More on Deductive Geometry
Important Terms
deductive geometry
congruent triangle
similar triangle
angle bisector
perpendicular bisector
median
concurrent
演繹幾何
全等三角形
相似三角形
角平分線
垂直平分線
中線
共點
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Name:____________(
) Class: F.3 (
P. 1
)
頂垂線 (高)
內心
內切圓
外心
外接圓
形心
垂心
altitude
in-centre
inscribed circle
circumcentre
circumcircle
centroid.
orthocenter
Revision Notes
1. Using the deductive approach to solve geometric problems that involve triangles.
Based on the properties or conditions of congruent triangles, similar triangles and isosceles triangles,
we can use deductive approach to prove and obtain more geometric results.
2. Some special lines in a triangle
(a) An angle bisector is a line segment which divides an angle into two equal parts.
B
e.g. In the figure, AD is the angle bisector of BAC .
D
A
C
(b) A perpendicular bisector is a line which bisects a side and is perpendicular to that side.
e.g. In the figure, DE is the perpendicular bisector of AC.
B
D
A
C
E
(c) A median is a line segment which joins a vertex to the mid-point of its opposite side.
e.g. In the figure, BD is a median.
B
A
C
D
(d) An altitude is a perpendicular line segment which drops from a vertex to its opposite side.
e.g. In the figure, BD is an altitude.
B
A
D
C
F.3 Mathematics Supplementary Notes
3.
Chapter 5 More on Deductive Geometry
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P. 2
Triangle Inequality
In any triangle, the sum of the lengths of any two sides is always longer than the length of the third
side.
e.g. In the figure, a + b > c,
a
b
b + c > a,
c + a > b.
c
4.
Relations between the lines in a triangle
(a) The 3 angle bisectors of a triangle are always concurrent.
The point of intersection is called the in-centre.
A circle (inscribed circle) can be constructed with
A
P
the in-centre as the centre and touching the 3 sides
of the triangle.
Q
I
e.g. In the figure, I is the in-centre
and PQR is the inscribed circle.
C
B
R
(b) The 3 perpendicular bisectors of a triangle are always concurrent.
The point of intersection is called the circumcentre.
A circle (circumcircle) can be constructed with
centre at the circumcentre and passing through
B
the 3 vertices of the triangle.
O
e.g. In the figure, O is the circumcentre
and ABC is the circumcircle.
A
C
(c) The 3 medians of a triangle are always concurrent.
The point of intersection is called the centroid.
e.g. In the figure, G is the centroid and
AG：GP =BG：GQ
=CG：GR
=2：1
B
R
P
G
A
C
Q
C
(d) The 3 altitudes of a triangle are always concurrent.
The point of intersection is called the orthocenter.
P
Q
H
e.g. In the figure, H is the orthocenter.
A
R
B
F.3 Mathematics Supplementary Notes
Chapter 5 More on Deductive Geometry
11/2005
P. 3
Theorem review:
Theorem The sum of the angles of a triangle is 1800.
i.e. a  b  c  1800
[  sum of  ]
Theorem In a triangle, the exterior angle is equal to the sum of the interior opposite angles,
i.e. c1  a  b
[ ext. of  ]
Theorem The sum of the interior angles of an n-sided convex polygon
is (n  2)  1800
[  sum of polygon ]
Theorem If the sides of a convex n-sided polygon are produced in order,
the sum of the exterior angles is 3600. [ sum of ext.  of polygon ]
A
Theorem The base angles of an isosceles triangle are equal,
i.e. when AB = AC, then ABC = ACB
[ base s, isos.  ]
A
Theorem When ABC = ACB , then AB = AC.
B
[ sides opp. equal s ]
B
Tests for Congruent Triangles
C
C
F.3 Mathematics Supplementary Notes
Chapter 5 More on Deductive Geometry
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P. 4
Tests for Similar Triangles
If
ABC = XYZ
BCA = YZX
CAB =ZXY
If
AB AC BC


k
XY XZ YX
CAB =ZXY
AB AC

k
XY XZ
If
A
A
c
b
b
c
B
B
C
C
a
X
X
kc
kb
kc
Y
Z
Y
[
A.A.A.
]
kb
[3 sides proportional]
then we have
ABC ～ XYZ
Exercise (Level I)
1. Find the unknowns of the following figures.
(a)
(b)
ka
[ ratio of 2 sides, inc. ]
Z
F.3 Mathematics Supplementary Notes
Chapter 5 More on Deductive Geometry
(c)
(d)
A
B
E
C
D
(e)
(f)
(g)
(h)
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P. 5
F.3 Mathematics Supplementary Notes
2.
Chapter 5 More on Deductive Geometry
In the figure, if a1 = b1、a2 = b2, prove that AC = BE.
11/2005
C
E
a1
D
a2 b2
A
3.
In the figure, prove that AB // CD.
4.
In the figure,
(a) prove that ABC  PQR;
(b) find QPR.
P. 6
b1
B
A
6
C
36.9
4.5
53.1
7.5
B
P
4
R
3
5
Q
F.3 Mathematics Supplementary Notes
Chapter 5 More on Deductive Geometry
11/2005
Level 2
5.
In the figure, AB = AC and AX = AY, prove that
(a)
(b)
Y
C
BCY  CBX;
BCZ is an isosceles triangle.
Z
A
X
B
6.
P. 7
In the figure, OX and OY are the perpendicular bisectors of AB and AC and OX = OY, show that
(a) OAX  OBX ;
(b) OB = OC.
C
Y
O
A
7.
X
B
In the figure, AD is the angle bisector of BAC, BAP is a straight line and AD//PC,
Show that AP = AC
C
D
B
A
P
F.3 Mathematics Supplementary Notes
8.
Chapter 5 More on Deductive Geometry
In the figure, AD  BC and BE  AC, prove
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CE BC

.
CD AC
A
E
B
9.
(a).
D
A
In the figure, M and N are the mid-points of AB and AC respectively.
Prove that AMN  ABC;
(b). Hence, prove that MN =
C
1
BC.
2
M
B
C
N
P. 8
F.3 Mathematics Supplementary Notes
Chapter 5 More on Deductive Geometry
11/2005
P. 9
**************************************************************************************
Level III (Optional)
1. In the figure, ACD and BGFC are two straight lines. AD // EG,
AC : CD = 2 : 1 and EF : FD = 2 : 3.
D
C
(a) Name two pairs of similar triangles and give reasons.
F
G
(b) Hence find the ratio AE : EB.
(Ans: (a) ABC ~ EBG (b)2：１)
A
B
E
2. In the figure, AB//DC, DA//CB and AD = BC. CD is produced to a point E. EB cuts DA and
CA at F and G respectively. Given that EF = 16 and FB = 8.
E
(a) Find the ratio of DF to CB.
(b) Prove that GCB  GAF, hence find the ratio of FG
to GB.
(Ans:(a)2:3 (b)1:3)
16
F
D
G
8
C
B
A
F.3 Mathematics Supplementary Notes
數學課外閱讀
Chapter 5 More on Deductive Geometry
11/2005
P. 10
<<典雅的幾何>> 作者 : 倫迪(Miranda Lundy) 、薩頓(Daud Sutton)
譯者 : 葉偉文
出版社: 天下遠見
內容簡介 :
一本讓人愛不釋手、想細心收藏的圖畫書！
從瓷磚、教堂窗戶到偉大的金字塔，這本書將許多建築的基本材料與其設
計相結合，引領我們體驗幾何之美。你可以從這本迷人的小冊子中看見三角形、
拱形、六角形及螺形的優美與單純，它們雖隨處可見卻蘊含著極大的學問。作
者之一的倫迪，在本書前半部（神聖的幾何）中展現二維度裡的空間，她用精
緻的手繪圖形來說明各形狀之間的關係、黃金分割的定義等，經由她的詮釋，
你將徹底改變以往對形狀的觀點。
五個柏拉圖立體和十三個阿基米得立體包括正方體、八面體、數不清到底
有幾面的二十面體等，而這十八種模型構成了三維空間的建材形狀，也是建築、
化學和原子物理的核心。同時身為是幾何學家與藝術家的作者薩頓，在本書第
二部——柏拉圖和阿基米得立體，為我們揭示這些立體之間所蘊含的純樸之美。
資料來源 : 商務網上書店
http://www.cp1897.com.hk/BookInfo?BookId=7535739636&SectionId=10&AllId=0&Action=1
F.3 Mathematics Supplementary Notes
Chapter 5 More on Deductive Geometry
11/2005
P. 11