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HKDSE Maths (Compulsory)

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HKDSE Maths (Compulsory) 中學文憑數學中學文憑數學中學文憑數學中學文憑數學(必修部份必修部份必修部份必修部份)

Ultimate Revision Handbook

終極溫習手冊終極溫習手冊終極溫習手冊終極溫習手冊

(中英對照版本)

DeductiveDeductiveDeductiveDeductive GeometryGeometryGeometryGeometry

演繹演繹演繹演繹幾何幾何幾何幾何

Includes more than 80 reasons essential for

HKDSE Maths (Compulsory) deductive geometry

包括八十多條中學文憑數學(必修部份)

演繹幾何理由

HKDSE Maths (Compulsory)

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1. Angles of Triangle (三角形的角三角形的角三角形的角三角形的角)

1.1 In any triangle,

在任何三角形內,

°=∠+∠+∠ 180CBA

Reason 理由:

[ ∠ sum of ∆ ] / [ ∆內角和]

1.2 In any triangle,

在任何三角形內,

ACDBACABC ∠=∠+∠

Reason 理由:

[ext. ∠ of ∆ ] / [ ∆的外角]

2. Angles and lines (角與線角與線角與線角與線)

2.1 If AO, BO, CO and DO intersect at O, then

若 AO、BO、CO 和 DO 交於 O,則

a + b + c + d = °360

Reason 理由:

[ ∠ s at a pt.] / [同頂角]

2.2 If AB and CD intersect at O, then

若 AB 和 CD 交於 O,則

a = b & x = y

Reason 理由:

[vert. opp. ∠ s] / [對頂角]

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3. Polygons (多邊形多邊形多邊形多邊形)

3.1 Sum of interior angles of a n-sided

polygon (n邊形內角和)

= °×− 180)2(n

Reason 理由:

[ ∠ sum of polygon] / [多邊形內角和]

3.2 Sum of exterior angles of convex polygon

凸多邊形外角和

= °360

Reason 理由:

[ext. ∠ of polygon] / [多邊形外角和]

4. Straight lines (直線直線直線直線)

4.1 If AOB is a straight line, then

若 AOB 是一條直線,則

a + b = °180

Reason 理由:

[adj. ∠ s on st. line] / [直線上的鄰角]

4.2 If a + b = °180 , then

若 a + b = °180 ,則

ΑΟΒ is a straight line.

AOB是一條直線。

Reason 理由:

[adj. ∠ s supp.] / [鄰角互補]

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5. Parallel lines (平行線平行線平行線平行線)

5.1 If PQ // RS, then (若 PQ // RS,則)

(a) a = b

Reason 理由:

[corr. ∠ s, PQ // RS] / [同位角,PQ // RS]

(b) b = c

Reason 理由:

[alt. ∠ s, PQ // RS] / [錯角,PQ // RS]

(c) b + d = °180

Reason 理由:

[int. ∠ s, PQ // RS] / [同旁內角,PQ // RS]

5.2 PQ // RS if either

若滿足以下任何一條件,則PQ // RS

(a) a = b

Reason 理由:

[corr. ∠ s eq.] / [同位角相等]

(b) b = c

Reason 理由:

[alt. ∠ s eq.] / [錯角相等]

(c) b + d = °180

Reason 理由:

[int. ∠ s supp.] / [同旁內角互補]

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6. Isosceles triangles (等腰三角形等腰三角形等腰三角形等腰三角形)

6.1 If AB = AC, then

若 AB = AC,則

CB ∠=∠

Reason 理由:

[base ∠ s, isos. ∆ ] / [等腰∆底角]

6.2 If CB ∠=∠ , then

若 CB ∠=∠ ,則

AB = AC

Reason 理由:

[side opp.. eq. ∠ s] / [等角對邊相等]

6.3 ABC∆ is an isosceles triangle with AB =

AC if

ABC∆ 是等腰三角形且 AB = AC 若

CB ∠=∠

Reason 理由:

[base ∠ s eq.] / [底角相等]

7. Equilateral triangle (等邊三角等邊三角等邊三角等邊三角形形形形)

7. If AB = BC = CA, then

若 AB = BC = CA,則

°=∠=∠=∠ 60CBA

Reason 理由:

[Property of equi. ∆ ] / [等邊∆性質]

A

B C

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8. Right-angled triangle (直角三角形直角三角形直角三角形直角三角形)

8.1 If °=∠ 90C , then

若 °=∠ 90C ,則

222

cba =+

Reason 理由:

[Pyth. theorem] / [畢氏定理]

8.2 If 222cba =+ , then

若 222cba =+ ,則

°=∠ 90C / ABC∆ is a right-angled ∆ .

Reason 理由:

[Converse of Pyth. theorem]

[畢氏定理的逆定理]

9. Congruent triangles (全等三角形全等三角形全等三角形全等三角形)

9.1 If 若 XYZABC ∆≅∆ , then 則

AB = XY

BC = YZ

AC = XZ

Reason 理由:

[corr. sides, ∆≅ s] / [ ∆≅ 的對應邊]

XA ∠=∠

YB ∠=∠

ZC ∠=∠

Reason 理由:

[corr. ∠ s, ∆≅ s] / [ ∆≅ 的對應角]

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9.2 XYZABC ∆≅∆ if 若

AB = XY

BC = YZ

CA = ZX

Reason 理由: [S.S.S.]

9.3 XYZABC ∆≅∆ if 若

YB ∠=∠

ZC ∠=∠

BC = YZ

Reason 理由: [A.S.A.]

9.4 XYZABC ∆≅∆ if 若

YB ∠=∠

ZC ∠=∠

BA = YX

Reason 理由: [A.A.S.]

9.5 XYZABC ∆≅∆ if 若

XA ∠=∠

BA = YX

AC = XZ

Reason 理由: [S.A.S.]

9.6 XYZABC ∆≅∆ if

XYZABC ∆≅∆ 若

BA = YX

AC = XZ

°=∠=∠ 90ZC

Reason 理由: [R.H.S.]

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10. Similar triangles (相似三角形相似三角形相似三角形相似三角形)

10.1 XYZABC ∆∆ ~ if 若

XA ∠=∠

YB ∠=∠

ZC ∠=∠

Reason 理由: [A.A.A.] / [等角]

10.2 XYZABC ∆∆ ~ if 若

ZX

CA

YZ

BC

XY

AB==

Reason 理由:

[3 sides prop.] / [三邊成比例]

10.3 XYZABC ∆∆ ~ if 若

ZX

CA

XY

AB= & XA ∠=∠

Reason 理由: [ratio of 2 sides, inc. ∠ ]

[兩邊成比例及夾角相等]

10.4 If 若 XYZABC ∆∆ ~ , then 則

ZX

CA

YZ

BC

XY

AB==

Reason 理由:

[corr. sides, ∆~ s] / [ ∆~ 的對應邊]

XA ∠=∠

YB ∠=∠

ZC ∠=∠

Reason 理由:

[corr. ∠ s, ∆~ s] / [ ∆~ 的對應角]

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11. Quadrilaterals 四邊形四邊形四邊形四邊形

11.1 Trapezium 梯形梯形梯形梯形

A quadrilateral having 1 pair of parallel sides

含有一對平行邊的四邊形

Properties 特性:

1. °=∠+∠ 180DA

2. °=∠+∠ 180CB

Reason 理由:

[Property of trapezium] / [梯形性質]

11.2 Parallelogram 平行四邊形平行四邊形平行四邊形平行四邊形

A quadrilateral having 2 pairs of parallel sides

含有一對平行邊的四邊形

1. AB // DC

2. AD // BC

Reason 理由:

[Property of //gram] / [平行四邊形性質]

Properties 特性特性特性特性:

11.2A If ABCD is a parallelogram, then

若為一平行四邊形,則

1. AB = CD

2. AD = BC

Reason 理由:

[opp. sides of //gram] / [平行四邊形對邊]

11.2B If ABCD is a parallelogram, then

若為一平行四邊形,則

1. CA ∠=∠

2. DB ∠=∠

Reason 理由:

[opp. ∠ s of //gram] / [平行四邊形對角]

11.2C If ABCD is a parallelogram, then

若為一平行四邊形,則

1. AO = OC

2. BO = OD

Reason 理由:

[diags. of //gram] / [平行四邊形對角線]

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11.3 Rhombus 菱形菱形菱形菱形

A parallelogram having 4 equal sides

含有4條等邊的平行四邊形

Properties 特性:

1. All properties of //gram

1. 所有平行四邊形的特性。

2. Diagonals bisect each interior angle

2. 對角線平分每隻內角。

3. Diagonals ⊥ to each other

3. 對角線互相垂直。

Reason 理由:

[Property of rhombus] / [菱形性質]

11.4 Rectangle 長方形長方形長方形長方形

A parallelogram having 4 right angles

含有4隻直角的平行四邊形

Properties 特性:

1. All properties of //gram

1. 所有平行四邊形的特性。

2. Diagonals are equal in length.

2. 對角線長度相等。

Reason 理由:

[Property of rectangle] / [長方形性質]

11.5 Square 正方形正方形正方形正方形

A parallelogram having 4 right angles & 4

equal sides

含有4隻直角及4條等邊的平行四邊形

Properties 特性:

1. All properties of //gram

1. 所有平行四邊形的特性。

2. Diagonals are equal in length.

2. 對角線長度相等。

3. Diagonals ⊥ to each other

3. 對角線互相垂直。

4. Angle between diagonal and side = °45

4. 對角線與邊之間的夾角為 °45

Reason 理由: [Property of square] / [正方形性質]

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11.6 Kite 鳶形鳶形鳶形鳶形

A quadrilateral having 2 pairs of adjacent equal

sides

含有2對相鄰等邊的四邊形

Properties 特性:

Diagonals ⊥ to each other

對角線互相垂直。

Reason 理由:

[Property of kite] / [鳶形性質]

12. Proofs of parallelogram (平行四邊形的證明平行四邊形的證明平行四邊形的證明平行四邊形的證明)

12.1 ABCD is a parallelogram if

ABCD為一平行四邊形若

1. AB = DC

2. AD = BC

Reason 理由: [opp. sides eq.] / [對邊相等]

12.2 ABCD is a parallelogram if

ABCD為一平行四邊形若

1. CA ∠=∠

2. DB ∠=∠

Reason 理由: [opp. ∠ s eq.] / [對邊角等]

12.3 ABCD is a parallelogram if

ABCD為一平行四邊形若

1. AO = OC

2. BO = OD

Reason 理由:

[diags. bisect each other] / [對角線互相平分]

12.4 ABCD is a parallelogram if

ABCD為一平行四邊形若

1. AB = DC

2. AB // DC

Reason 理由:

[opp. sides eq. and //] / [一組對邊相等且平行]

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13. Mid-point theorem (中點定理中點定理中點定理中點定理)

13.1 If AM = MB and AN = NC, then

若AM = MB及AN = NC,則

1. MN // BC

2. BCMN2

1=

Reason 理由: [Mid-pt. thm.] / [中點定理]

13.2 If MN // BC and BCMN2

1= , then

若MN // BC及 BCMN2

1= ,則

1. AM = MB

2. AN = NC

Reason 理由:

[Converse of mid-pt. thm.] / [中點逆定理]

14. Intercept theorem (截線定理截線定理截線定理截線定理)

14.1 If AB // CD // EF, then

若AB // CD // EF,則

CE

AC

DF

BD=

Reason 理由: [Intercept thm.] / [截線定理]

14.2 If CE

AC

DF

BD= , then / 若

CE

AC

DF

BD= ,則

AB // CD // EF

Reason 理由:

[Converse of intercept thm.] / [截線逆定理]

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15. Relationships between sides and angles of triangle

(三角形三角形三角形三角形的邊及角的關係的邊及角的關係的邊及角的關係的邊及角的關係)

15.1. Triangle inequality 三角形不等式

In any triangle 在任何三角形內

a + b > c

a + c > b

b + c > a

Reason 理由:

[Triangle inequality] / [三角形不等式]

15.2. Greater angle, greater side 大角對大邊

A. If CBA ∠>∠>∠ , then

若 CBA ∠>∠>∠ ,則

a > b > c

Reason 理由:

[Greater ∠ , greater side] / [較大∠,較大邊]

B. If a > b > c, then

若 a > b > c,則

CBA ∠>∠>∠

Reason 理由:

[Greater side, greater ∠ ] / [較大邊,較大∠ ]

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16. Points of concurrence in triangle (三角形中的共點三角形中的共點三角形中的共點三角形中的共點)

16.1 Centroid 形心

If point G is the centroid of the triangle, then

若點 G為三角形的形心,則

CD = DB

Reason 理由:

[Centroid of ∆ ] / [ ∆的形心]

16.2 Incentre 內心

If point I is the incentre of the triangle, then

若點 I為三角形的內心,則

BADCAD ∠=∠

Reason 理由:

[Incentre of ∆ ] / [ ∆的內心]

16.3 Orthocentre 垂心

If point R is the orthocentre of the triangle,

then

若點 R為三角形的垂心,則

CBAD ⊥

Reason 理由:

[Orthocentre of ∆ ] / [ ∆的垂心]

16.4 Circumcentre 外心

If point C is the circumcentre of the triangle,

then

若點 C為三角形的外心,則

DBCE ⊥ & DE = EB

Reason 理由:

[Circumcentre of ∆ ] / [ ∆的外心]

A

B

C

G

D

A

B

C

I

D

A

B

C

R

D

A

B

D

C

E

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17. Chords of a circle (圓內的弦圓內的弦圓內的弦圓內的弦)

17.1. A perpendicular line from the centre of a

circle to a chord bisects the chord.

由圓心畫一條垂線至任何一條弦會平分該

弦。

If ON ⊥ AB, then AN = NB.

若ON ⊥ AB,則AN = NB。

Reason 理由:

[ ⊥ from centre to chord bisects chord]

[圓心至弦的垂線平分弦]

17.2. The line joining the centre of a circle

and the mid-point of a chord is perpendicular

to the chord.

由圓心畫一條直線至某弦的中點,則該直線

必與孤互相垂直。

If AM = MB, then OM ⊥ AB.

若AM = MB,則OM ⊥ AB。

Reason 理由:

[line joining centre and mid-pt. of chord

⊥ chord] / [圓心至弦中點的連線垂直弦]

17.3 The centre of a circle lies on the

perpendicular bisector of a chord.

圓心為於弦的垂直平分線上。

Reason 理由:

[ ⊥ bisector of chord passes through centre]

[弦的垂直平分線穿過圓心]

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17.4 If two chords of a circle are equal in

length, then they are equidistant from the

centre.

若某圓內的兩條弦相等,則該兩條弦與圓心

的距離相等。

If AB = CD, then OM = ON.

若AB = CD,則OM = ON。

Reason 理由:

[eq. chords equidistant from centre]

[等弦與圓心等距]

17.5 If two chords of a circle are equidistant

from the centre, then their lengths are

equal.

若兩條弦與圓心等距,則該兩條弦的長度相

等。

If OM = ON, then AB = CD.

若OM = ON,則AB = CD。

Reason 理由:

[chords equidistant from centre eq.]

[與圓心等距的弦等長]

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18. Angles in a circle (圓內的角圓內的角圓內的角圓內的角)

18.1 The angle at the centre of a circle

subtended by an arc is twice the angle at the

circumference subtended by the same arc.

一個弧所對的圓心角是該弧所對的圓周角的

兩倍。

x = 2y.

Reason 理由:

[ ∠ at centre twice ∠ at ce

]

[圓心角兩倍於圓周角]

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18.2 If AB is a diameter and P is any point on

the circumference except A and B, then

若 AB 是直徑,而 P 是圓周上除 A 和 B

以外的任意一點,則

∠ APB = °90 .

Reason 理由:

[ ∠ in semi-circle] / [半圓上的圓周角]

18.3 If ∠ APB = °90 , then AB is a diameter.

若∠ APB = °90 ,則 AB為直徑。

Reason 理由:

[Converse of ∠ in semi-circle]

[半圓上的圓周角逆定理]

18.4 Angles in the same segment of a circle are equal.

同弓形內的圓周角相等。

x = y

Reason 理由: [ ∠ s in the same segment] / [同弓形內的圓周角]

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19. Relationships among Arcs, Chords & Angles (弧弧弧弧,,,,弦及角之間的關係弦及角之間的關係弦及角之間的關係弦及角之間的關係)

19.1 Equal angles at the centre of a circle (or

equal circles) stand on equal arcs and equal

chords.

在同一個圓或等圓上,相等的圓心角所對應

的弧及弦相等。

If a = b, then / 若 a = b,則

1. ∩

AB = ∩

CD

[eq. ∠ s, eq. arcs] / [等角對等弧]

2. AB = CD

[eq. ∠ s, eq. chords] / [等角對等弦]

19.2 Equal chords on a circle stand on equal

arcs and equal angles.

在同一個圓或等圓上,相等的弦所對應的圓

心角及弧相等。

If AB = CD, then / 若 AB = CD,則

a = b

[eq. chords, eq. ∠ s] / [等弦對等角]

AB = ∩

CD

[eq. chords, eq. arcs] / [等弦對等弧]

19.3 Equal arcs in a circle (or equal circles)

subtend equal angles at the centre.

在同一個圓或等圓上,相等的弧所對應的圓

心角及弦相等。

If ∩

AB = ∩

CD , then / 若∩

AB = ∩

CD,則

a = b

[eq. arcs, eq. ∠ s] / [等弧對等角]

AB = CD

[eq. arcs, eq. chords] / [等弧對等弦]

HKDSE Maths (Compulsory)

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19.4 In a circle (or equal circles), the lengths of

arcs are proportional to the sizes of angles at

the centre subtended by the arcs.

在同一個圓或等圓上,弧長與所對的圓心角

成比例。

AB : ∩

BC = m : n

Reason 理由:

[arcs prop. to ∠ s at centre]

[弧長與圓心角成比例]

19.5 In a circle (or equal circles), the lengths of

arcs are proportional to the sizes of angles at

the circumference.

在同一個圓或等圓上,弧長與所對的圓周角

成比例。

AB : ∩

BC = x : y

Reason 理由:

[arcs prop. to ∠ s at ce

]

[弧長與圓周角成比例]

Vinci 哥哥哥哥哥哥哥哥背景背景背景背景::::

• 2010年中學會考數學科學生成績:1A,4B,3C

• 2010 (Nov) IGCSE Maths成績:1位同學於 Edexcel奪得 A*級,

2同學於 Cambridge奪得 A級。

HKDSE Maths (Compulsory)

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20. Cyclic Quadrilateral (圓內接四邊形圓內接四邊形圓內接四邊形圓內接四邊形)

20.1 The opposite angles of a cyclic

quadrilateral are supplementary.

圓內接四邊形的兩個對角互補。

°=+ 180yx

Reason 理由:

[opp. ∠ s, cyclic quad.]

[圓內接四邊形對角]

20.2 For a cyclic quadrilateral, an exterior

angle is equal to its interior opposite angle.

圓內接四邊形任何一個外角與其內對角相

等。

a = b

Reason 理由:

[ext. ∠ s, cyclic quad.]

[圓內接四邊形外角]

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HKDSE Maths (Compulsory)

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21. Tests for Concyclic Points (共圓點的測試共圓點的測試共圓點的測試共圓點的測試)

21.1 If p = q, then A, B, Q and P are concyclic.

若 p = q,則 A、B、C 和 D 共圓。

Reason 理由:

[Converse of ∠ s in the same segment]

[同弓形內的圓周角的逆定理]

21.2 If °=∠+∠ 180CA or °=∠+∠ 180DB ,

then A, B, C and D are concyclic.

若 °=∠+∠ 180CA ,則 A、B、C 和 D 共

圓。

Reason:

[opp. ∠ s supp.] / [對角互補]

21.3 If p = q, then A, B, C and D are concyclic.

若 p = q,則 A、B、C 和 D 共圓。

Reason 理由:

[ext. ∠ = int. opp. ∠ ]

[外角等於內對角]

HKDSE Maths (Compulsory)

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22. Properties of Tangents of a circle (圓的切線的特性圓的切線的特性圓的切線的特性圓的切線的特性)

22.1 If PQ is a tangent to the circle at T, then

若 PQ 為圓在 T 點的切線,則

OT ⊥ PQ.

Reason 理由:

[tangent ⊥ radius] / [切線 ⊥ 半徑]

Conversely, if OT ⊥ PQ, then PQ is a tangent

to the circle at T.

反過來說,若OT ⊥ PQ,則 PQ 為圓在 T

點的切線。

Reason 理由:

[converse of tangent ⊥ radius]

[切線⊥半徑的逆定理]

22.2 Perpendicular to a tangent at its point of

contact passes through the centre of the circle.

切線的垂線於接觸點穿過圓心。

22.3 If TP and TQ are two tangents to a circle

at P and Q respectively, then

若從圓外的一點 T 分別作兩條與圓切於 P

和 Q 的切線 TP 和 TQ,則

(a) TP = TQ;

(b) ∠ TOP = ∠ TOQ;

(c) ∠ PTO = ∠ QTO.

Reason 理由:

[tangents from ext. pt.] / [tangent properties]

[由外點引切線] / [切線性質]

HKDSE Maths (Compulsory)

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23. Angle in alternate segment (交錯弓形的圓周角交錯弓形的圓周角交錯弓形的圓周角交錯弓形的圓周角)

23.1 If PQ is a tangent to the circle at A, then

若 PQ 為圓在 A 點的切線而 AC 為弦,則

x = y

Reason 理由:

[ ∠ in alt. segment] / [交錯弓形的圓周角]

23.2 If x = y, then PQ is the tangent to the

circle at A.

若 x = y,則 PQ 為該圓在 A 點的切線。

Reason 理由:

[Converse of ∠ in alt. segment]

[交錯弓形的圓周角的逆定理]

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