basic properties of circles (ii) · 7 basic properties of circles (ii) 3 5. in the figure, oab is a...

1

Basic Properties of Circles (II)

圓的基本特性 (二)

Exercises(練習)

1. In the figure, AB is a diameter of the circle, DC is the tangent to the circle at D

and BAD = 32. If ABC is a straight line, find x.

在圖中，AB 是圓的一條直徑，DC 是該圓於 D 的切線，而 BAD = 32。

若 ABC 是一條直線，求 x。

Join OD.

64

322

2 DABOBQ

( at centre twice at ☉ce)

In △ODC,

90ODC (tangent radius)

26

1809064

180

x

x

OCDODCDOC

( sum of △)

2. In the figure, CB and CA are tangents to the circle at B and D respectively.

AEOB and BFD are straight lines and BAC = 36. Find

(a) AOC,

(b) DBC.

在圖中，CB 和 CA 分別是圓於 B 和 D 的切線。若 AEOB 和 BFD 都是

直線，而 BAC = 36，求

(a) AOC；

(b) DBC。

(a) 90OBC (tangent radius)

In △ABC,

54

1809036

180

ACB

ACB

ACBABCBAC ( sum of △)

DCOOCB (tangent properties)

Question Bank

2

∴

27

2

54OCB

117

9027

OBCOCBAOC

(ext. of △)

(b) CBCD (tangent properties)

BDCDBC (base s, isos. △)

In △DBC,

180BDCDBCBCD ( sum of △)

63

2

54180

180254

DBC

DBC

3. In the figure, AB is a diameter of the circle, PQ is the tangent to the circle at B and

AOC = 70. Find x.

在圖中，AB 是圓的一條直徑，PQ 與圓相切於 B，而 AOC = 70。求 x。

35

702

2

OBC

OBC

AOCOBC ( at centre twice at ☉ce)

∵ 90OBQ (tangent radius)

∴

55

9035

90

x

x

CBQOBC

4. In the figure, AC = 12 cm, OC = 9 cm and CB = 16 cm. Show that AB is the

tangent to the circle at A.

在圖中，AC = 12 cm，OC = 9 cm 及 CB = 16 cm。證明 AB 與該圓相切於 A。

In △AOC,

2

222

222

cm 225

cm)912(

OCACAO

Pyth. theorem

In △ACB,

2

222

222

cm 400

cm)1612(

CBACAB

Pyth. theorem

2

222

cm 625

cm)400225(

AOAB

2

222

cm 625

cm)169(

OB

∴ 222 OBAOAB

∴ ABOA converse of Pyth. theorem

∴ AB is the tangent to the circle at A. converse of tangentradius

7 Basic properties of Circles (II)

3

5. In the figure, OAB is a straight line, AB = AC and ABC = 30. Prove that BC is

the tangent to the circle at C.

在圖中，OAB 是一條直線，AB = AC 及 ABC = 30。證明 BC 與該圓相切於

C。

CABA given

ABCACB base s, isos. △

30

60

3030

ACBABCOAC

ext. of △

COAO radii

60

OACOCA

base s, isos. △

∵

90

6030

OCAACBOCB

∴ BC is the tangent to the circle at C. converse of tangent radius

6. In the figure, AB, BC, CD and DA are tangents to the circle at P, Q, R and S

respectively. If AB = 10 cm, BC = 8 cm and CD = 6 cm, find DA.

在圖中，AB、BC、CD 和 DA 分別是圓於 P、Q、R 和 S 的切線。若 AB = 10 cm，

BC = 8 cm 及 CD = 6 cm，求 DA。

Let cm xPB .

cm )10( x

PBABAP

cm )10( x

APAS

(tangent properties)

cm x

PBBQ


cm )8( x

BQBCQC

cm )8( x

QCRC


cm )2(

cm )]8(6[

x

x

RCDCDR

cm )2(

x

DRDS


cm 8

cm )]2()10[(

xx

DSASDA

7. In the figure, DB and DC are tangents to the circle at B and C respectively. ABE

and EDC are straight lines. If BAC = 30 and BED = 50, find x.

在圖中，DB 和 DC 分別是圓於 B 和 C 的切線。ABE 和 EDC 都是直線。

若 BAC = 30 及 BED = 50，求 x。

30

BACBCD

( in alt. segment)

Question Bank

4

30

BACCBD

( in alt. segment)

In △ECB,

70

180)30(3050

180

x

x

CBEECBBEC ( sum of △)

8. In the figure, BA and BC are tangents to the circle at A and C respectively.

Prove that OB is the perpendicular bisector of AC.

在圖中，BA 和 BC 分別是圓於 A 和 C 的切線。證明 OB 是 AC 的垂直

平分線。

BCAB tangent properties

KBCABK tangent properties

BKBK common side

∴ △ABK△CBK SAS

KCAK corr. sides, △s

BKCAKB corr. s, △s

180BKCAKB adj. s on st. line

∴ 90BKCAKB

∴ OB is the perpendicular bisector

of AC.

9. In the figure, AFD and BEC are straight lines. Determine whether

the following statements are true.

(a) A, B, E and F are concyclic.

(b) F, E, C and D are concyclic.

(c) A, B, C and D are concyclic.

在圖中，AFD 和 BEC 都是直線。判斷下列各句子是否正確。

(a) A、B、E 和 F 共圓。

(b) F、E、C 和 D 共圓。

(c) A、B、C 和 D 共圓。

(a) ∵ 110BAF and 80FEC

FECBAF

∴ A, B, E and F are not concyclic.

(b) ∵

180

190

80110

FECFDC

∴ F, E, C and D are not concyclic.

(c) ∵ 20ADB and 20ACB

∴ ACBADB

∴ A, B, C and D are concyclic. (converse of s in the same segment)


5

10. In the figure, ED is the tangent to the circle at C and DBA is a straight line.

(a) Prove that △BCD △CAD.

(b) If BD = 4 cm and CD = 6 cm, find AB.

在圖中，ED 是圓於 C 的切線，而 DBA 是一條直線。

(a) 證明 △BCD △CAD。

(b) 若 BD = 4 cm 及 CD = 6 cm，求 AB。

(a) Consider △BCD and △CAD.

CADBCD in alt. segment

CDABDC common angle

∴ △BCD△CAD AAA

(b) ∵ △BCD △CAD

∴

cm 2

36cm 4

cm4

cm 6

3

2

cm 4

cm 6

6

4

AB

AB

AB

AD

CD

CD

BD

∴ cm 5AB

11. In the figure, PQ is the tangent to the circle at Q. PAMB is a straight line and

M is the mid-point of chord AB. Prove that O, M, P and Q are concyclic.

在圖中，PQ 是圓於 Q 的切線。PAMB 是一條直線，而 M 是弦 AB 的中

點。證明 O、M、P 和 Q 共圓。

90OMA line joining centre to mid-pt. of chord chord

90OQP tangent radius

180

9090

OQPOMP

∴ O, M, P and Q are concyclic. opp. s supp.

12. In the figure, ABE and EKT are circles with centre O and D

respectively. The two circles touch each other externally at E. CE is their

common tangent at E. ARB, OSB, AOED, RSET and BCKDT are straight

lines. It

is given that 1:2:

BEAB and BTS = 15.

(a) Prove that BCKDT is the tangent to the circle ABE at B.

(b) Prove that the circles ABE and EKT have the same radii.

(c) Prove that △OBD △CED.

在圖中，ABE 和 EKT 兩個圓的圓心分別為 O 和 D。CE 與該兩圓同時相切於 E。ARB、OSB、

AOED、RSET 和 BCKDT 都是直線。

已知 1:2:

BEAB 及 BTS = 15。

(a) 證明 BCKDT 是圓 ABE 於 B 的切線。

Question Bank

6

(b) 證明 ABE 和 EKT 兩個圓半徑的長度相等。

(c) 證明 △OBD △CED。

(a) ETKEDK 2 at centre twice at ☉ce

30

152

BE

AB

BOE

AOB

arcs prop. to s at centre

1

2

∴ BOEAOB 2

60

1803

180

BOE

BOE

BOEAOB

adj. s on st. line

In △OBD,

1803060

180

OBD

EDKOBDBOE sum of △

∴ 90OBD

∴ BCKDT is the tangent to converse of tangent

the circle at ABE at B. radius

(b) In △OBD,

90OBD from (a)

60BOE from (a)

EDOEOB

EDOE

OB

EDOE

OB

EDOE

OBBOE

2

2

1

60cos

cos

∵ OEOB radii

∴

EDOB

EDOBOB

2

∴ Circles ABE and EKT have the

same radii.

(c) ODBEDC common angle

90OBD tangent radius

90CED tangent radius

∴ OBDCED

∴ △OBD ~ △CED AAA

13. In the figure, ABC and DEF are straight lines and DC // FA. Prove that B, C,

D and E are concyclic.

在圖中，ABC 和 DEF 都是直線，而 DC // FA。證明 B、C、D 和 E 共

圓。

BEDBAF ext. , cyclic quad.

180BCDBAF int. s, FA // CD

∴ 180BCDBED

∴ B, C, D and E are concyclic. opp. s supp.


7

14. In the figure, BCD and BFE are straight lines, AB = AD and CB = CA.

Prove that ABCF is a cyclic quadrilateral.

在圖中，BCD 和 BFE 都是直線，AB = AD 及 CB = CA。證明 ABCF 是

一個圓內接四邊形。

ADAB given

∴ ADBABD base s, isos. △

CABC given

∴ CABCBA base s, isos. △

∴ ADBCAB

BFCADB ext. , cyclic quad.

∴ BFCCAB

∴ ABCF is a cyclic quadrilateral. converse of s in the same segment

15. In the figure, ABD and BCD are two equal circles with centre O and O’

respectively. They intersect at B and D. The centre of each of these circles lies

on the other circle. AOO’C and BD intersect at K.

(a) Prove that AB and AD are tangents to the circle BCD at B and D

respectively.

(b) Prove that CB and CD are tangents to the circle ABD at B and D

respectively.

(c) Prove that △O’AD △OCD.

(d) Prove that ABCD is a rhombus.

在圖中，ABD 和 BCD 兩個等圓的圓心分別為 O 和 O’。該兩圓相交於 B 和 D，而它們的圓心則

分別位於另一個圓上。AOO’C 與 BD 相交於 K。

(a) 證明 AB 和 AD 分別是圓 BCD 於 B 和 D 的切線。

(b) 證明 CB 和 CD 分別是圓 ABD 於 B 和 D 的切線。

(c) 證明 △O’AD △OCD。

(d) 證明 ABCD 是一個菱形。

(a) 90OAB in semi-circle

90OAD in semi-circle

∴ AB and AD are tangents converse of

to the circle BCD at B and D tangent radius

respectively.

(b) 90OBC in semi-circle

90ODC in semi-circle

∴ CB and CD are tangents converse of

to the circle ABD at B and D tangent radius

respectively.

(c) ∵ ABD and BCD are equal

circles with centre O and O

respectively.

∴ DOOD radii

∵ DOOC 2 and ODAO 2

∴ AOOC

90ODCDAO in semi-circle

∴ △ ADO △OCD RHS

Question Bank

8

(d) ∵ AB and AD are tangents proved in (a)

to circle BCD at B and D

respectively.

∴ ADAB tangent properties

∵ CB and CD are tangents to proved in (a)

circle ABD at B and D

respectively.

∴ CDCB tangent properties

∵ △ ADO △OCD proved in (c)

∴ AD = CD corr. sides, △s

∴ CBCDADAB

∴ ABCD is a rhombus.

16. In the figure, ABCD is a semi-circle with centre O. AB is the tangent

to the circle OBC at B. MAD, MBN and NCD are straight lines. It is

given that OBC = x, MBN NCD and BC // MAD.

(a) Prove that MAD is the tangent to the circle OBC at O.

(b) Find x.

(c) Prove that NCD is the tangent to the circle OBC at C.

(d) Prove that MBN is the tangent to the semi-circle ABCD at B.

在圖中，半圓 ABCD 的圓心是 O。AB 是圓 OBC 於 B 的切線。

MAD、MBN 和 NCD 都是直線。已知 OBC = x，MBN NCD 和

BC // MAD。

(a) 證明 MAD 是圓 OBC 於 O 的切線。

(b) 求 x。

(c) 證明 NCD 是圓 OBC 於 C 的切線。

(d) 證明 MBN 是半圓 ABCD 於 B 的切線。

(a) OBOC radii

xOBCOCB base s, isos. △

x

OBCBOM

alt. s, MD // BC

∴ BOMOCB

∴ MAD is the tangent to the converse of in

circle OBC at O. alt. segment

(b) Consider △AOB.

x

OBCBOA

(alt. s, MD // BC)

OAOB (radii)

OABOBA (base s, isos. △)

x

OCBOBA

( in alt. segment)

∴ BOA = OBA = OAB = x

60

1803

180

x

x

OABOBABOA

( sum of △)

(c)

60

x

CBOCOD

in alt. segment


9

Consider △OCD.

ODOC radii

ODCOCD base s, isos. △

60

180602

180

OCD

OCD

CODODCOCD

sum of △

∴ 60OCDCBO

∴ NCD is the tangent to the converse of in alt.

circle OBC at C. segment

(d) In OCD,

60

OCDODC

from (c)

In △MDN,

30

1809060

180

DMN

DMN

MNDMDNDMN

sum of △

In △MOB,

60

xBOM

from (a)

90

1803060

180

MBO

MBO

BMOMBOBOM

sum of △

∴ MBN is the tangent to the converse of tangent

semi-circle ABCD at B. radius

17. In the figure, AD is a diameter of the circle with centre O. KCMN is

the tangent to the circle at C. BQPF, FSRC, AQORDM and APSEN are

straight lines. It is given that ANM = 30, RCM = 75,

1:3:

DEAD and BC // AD.

(a) Find ADE, PQR and BFC.

(b) Determine whether the following points are concyclic.

(i) D, M, N and E

(ii) S, R, D and E

(iii) P, Q, R and S

(iv) B, C, R and Q

在圖中，AD 是圓的一條直徑，而圓心是 O。KCMN 是圓於 C 的

切線。BQPF、FSRC、AQORDM 和 APSEN 都是直線。已知

ANM = 30，RCM = 75， 1:3:

DEAD 及 BC // AD。

(a) 求 ADE、PQR 和 BFC。

(b) 判斷下列各點是否共圓。

(i) D、M、N 和 E

(ii) S、R、D 和 E

(iii) P、Q、R 和 S

(iv) B、C、R 和 Q

(a) 90AED ( in semi-circle)

Question Bank

10

∴

30

90

1

3

DAE

DAE

DAE

AED

DE

AD

(arcs prop. to s at ☉ce)

In △ADE,

180AEDDAEADE ( sum of △)

60

1809030

ADE

ADE

75

FCMFBC

( in alt. segment)

75

FBCPQR

(corr. s, AM // BC)

In △AMN,

60

3030

ANBMANAMK

60

AMKBCK

(corr. s, BC // AM)

60

BCKBFC

( in alt. segment)

(b) (i) From (a), 60ADE

∵ ENMADE

∴ D, M, N and E are not concyclic.

(ii) From (a), 75PQRFQR

and 60BFCQFR

In △FQR,

135

6075

QFRFQRCRM

(ext. of △)

180

225

90135

SEDSRM

∴ S, R, D and E are not concyclic.

(iii) 75PQR (from (a))

In △SCN,

75

3075180CSN

( sum of △)

∴ CSNPQR

∴ P, Q, R and S are concyclic. (ext. = int. opp. )

(iv) 75FBC (from (a))

∴ 75QBC

135

SRMQRC

(vert. opp. s)

∵

210

13575QRCQBC

∴ B, C, R and Q are not concyclic.


11

18. In the figure, ACE and AEG are two circles with centre O and O’

respectively. They intersect at A and E. ABC is the tangent to the circle

AEG at A. AD and BE intersect at O. CDEF and FGO’A are straight lines. It

is given that AFC = 30.

(a) Find AOE.

(b) Prove that CDEF is the tangent to the circle AEG at E.

(c) Determine whether the following points are concyclic.

(i) O, B, C and D

(ii) A, B, D and E

在圖中，ACE 和 AEG 兩個圓的圓心分別是 O 和 O’。該兩圓相交於

A 和 E。ABC 是圓 AEG 於 A 的切線。AD 與 BE 相交於 O。CDEF

和 FGO’A 都是直線，而 AFC = 30。

(a) 求 AOE。

(b) 證明 CDEF 是圓 AEG 於 E 的切線。

(c) 判斷下列各點是否共圓。

(i) O、B、C 和 D

(ii) A、B、D 和 E

(a) 90CAF (tangent radius)

180AFCCAFACF ( sum of △)

60

1803090

ACF

ACF

ACEAOE 2 ( at centre twice at ☉ce)

602

120

(b)

Join EO .

60ACF proved in (a)

∴

60

ACFFOE

ext. , cyclic quad.

In △ OEF ,

90

1803060

180

EFO

EFO

OEFFOEEFO

sum of △

∴ CDEF is the tangent to the converse of tangent

circle AEG at E. radius

(c) (i) From (a), 60ACF

∴ 60BCD

Question Bank

12

120

AOEBOD

vert. opp. s

∵

180

12060

BODBCD

∴ O, B, C and D are concyclic. opp. s supp.

(ii)

Join AE.

Consider △AOE.

OEOA radii

OEAOAE base s, isos. △

CADOEA in alt. segment

BECOAE in alt. segment

∴ BECCAD

∴ A, B, D and E are concyclic. converse of s

in the same

segment


13

Pre-requisite Questions

預備測驗

1. Find the unknowns in the following figures.

求下列各圖中的未知量。

(a)

(b)

(a)

Draw GEH and IFJ such that AB // GH // IJ // CD.

134

46180

180 ABEBEG (int. s, GH // AB)

153

134287

287 BEGFEG

153

FEGEFJ (alt. s, IJ // GH)

137

43180

180 DCFCFJ

(int. s, IJ // CD)

∴

290

137153

CFJEFJx

(b)

56

180124

180

x

x

BADx (int. s, BC // AD)

124

BADy (corr. s,

2. In the figure, AC // DE, 45,64 CBEABD , DBE and BFC

are straight lines. Find

在圖中，AC // DE， 45,64 CBEABD ，DBE 和 BFC 都

是直線。求

(a) x,

Question Bank

14

(b) y.

(a) FBEABDABF 180

(adj. s on st. line)

71

4564180

71

ABFx

(base s, isos. △ )

(b)

45

CBEACB

(alt. s, AC // DE)

In △ AFC,

26

7145

y

y

xACFy

(ext. of

3. In the figure, QM // PN, 30,100 PQRMQP , QSR and

NPR are straight lines. Find x.

在圖中，QM // PN， 30,100 PQRMQP ，QSR 和 NPR

都是直線。求 x。

50

180130

180)30100(

180

QRN

QRN

QRN

MQRQRN (int. s, RN // QM)

In △ PSR,

50

250

x

xx

PSQPRSRPS

(ext. of △ )

4. Find the unknown in each of the following figures.

求下列各圖中的未知量。

5.

(a)

(b)

(a) 90QRP ( in semi-circle)

15

906

180905

180

y

y

yy

QRPQPRPQR ( sum of △ )

(b)

260

1302

2Reflex RQPROP



15

100

260360

reflex 360 ROPx (s at a pt.)

5. In the figure, AB // CD, AE and CE bisect ∠BAC and ∠DCA

respectively. Find reflex∠AEC.

在圖中，AB // CD，AE 和 CE 分別平分 ∠BAC 和 ∠DCA。求

優角 AEC。

With the notations in the figure,

aEACBAE

and bECDACE .

90

18022

180

ba

ba

ACDBAC (int. s, AB // CD)

∴

90

180 baAEC

( sum of △ )

∴

270

90360

360Reflex AECAEC (s at a pt.)

6. In the figure, a rectangle PQRS of length 32 cm is inscribed in a circle. If the

radius of the circle is 20 cm, find the perimeter of the rectangle PQRS.

在圖中，一個長度為 32 cm 的長方形 PQRS 內接於圓。若圓的半徑為

20 cm，求長方形 PQRS 的周界。

Draw a line MN passes through O and perpendicular to PS.

PMMS (line from centre chord bisects chord)

∴ cm 16MS

In △ OSM,

cm 12

cm1620 22

22

MSOSOM

(Pyth. theorem)

QRPS (property of rectangle)

∴ ONOM (equal chords, equidistant from centre)

∴ cm 12 OMON

∴ cm 24MN

Question Bank

16

∴ cm 24SR

The perimeter of

cm 112

cm )2432(2

PQRS

7. In the figure, PR is a chord of the circle, 27QPR and 40QRP .

Find

(a) ∠PQR,

(b) ∠POR.

在圖中，PR 為圓上的弦， 27QPR 及 40QRP 。求

(a) ∠PQR；

(b) ∠POR。

(a) In △ PQR,

113

4027180

180 QRPQPRPQR ( sum of △ )

(b)

226

1132

2Reflex PQRPOR


134

226360

reflex 360 PORPOR

8. In the figure, the radius of the circle is 10 cm. ON intersects AB at M and

ON ⊥ AB. If MN = 4 cm, find AB.

在圖中，圓的半徑為 10 cm。ON 與 AB 相交於 M，且 ON ⊥ AB。若 MN =

4 cm，求 AB。

Join OB.

cm 6

cm )4 10(

MNONMO

In △ OBM,

cm 8

cm610 22

22

MOOBMB

(Pyth. theorem)

cm 8

MBAM (line from centre chord bisects chord)


17

∴

cm 16

MBAMAB

9. In the figure, PR intersects QS at M and 117QMR . Find the value

of x.

在圖中，PR 與 QS 相交於 M，且 117QMR 。求 x。

23

x

QSRQPR

(s in the same segment)

In △ PQM,

20

1005

117175

117)23()152(

x

x

x

xx

QMRQPMPQM

(ext. of △ )

10. In the figure, AB is a diameter of the circle, 40DBC and

22DCA . Find

(a) ∠CBA,

(b) ∠CAB.

在圖中，AB 是圓的一條直徑， 40DBC 及 22DCA 。求

(a) ∠CBA；

(b) ∠CAB。

(a)

22

DCADBA


62

2240

DBACBDCBA

(b)

09 ACB ( in semi-circle)

In △ ABC,

28

6290180

801 CBAACBCBA

( sum

11. In the figure, 6:4:3:: CBA , find ∠D.

在圖中， 6:4:3:: CBA ，求 ∠D。

∵ 6:4:3:: CBA

∴ tA 3 , tB 4 and tC 6 , where 0t

20

1809

18063

180

t

t

tt

CA

(opp. s, cyclic quad.)

∴

80

204B

Question Bank

18

100

18080

180

D

DB


12. In the figure, BCE is a straight line and 50DCE . Find ∠BOD.

在圖中，BCE 是一條直線及 50DCE 。求 ∠BOD。

50

DCEBAD (ext. , cyclic quad.)

100

502

2 BADBOD


13. In the figure, 65PSQ and 25OPS , find

(a) ∠OPQ,

(b) QRS .

在圖中， 65PSQ 及 25OPS ，求

(a)∠OPQ；

(b) QRS 。

(a)

130

652

2 PSQPOQ


OQOP (radius)

OQPOPQ (base s, isos. △ )

180OQPOPQPOQ ( sum of △ )

25

2

130180

1802130

OPA

OPQ

(b) ∵

50

2525

OPQSPOQPS

130

18050

180

QRS

QRS

QPSQRS



19

Level 1 Questions

程度 1 題目

1. In the figure, AP is the tangent to the circle at P. Find x.

在圖中，AP 是圓於 P 的切線。求 x。

OPA = 90 (tangent radius)

OB = OP (radius)

= 7 cm

(rejected)32or18

0)32)(18(

057614

6251449

)7(247

2

2

222

222 theorem)Py th.(

xx

xx

xx

xx

x

OAPAOP

2. In the figure, AP is the tangent to the circle at P. AOB is a straight line

and APBP . Find ∠POA.

在圖中，AP 是圓於 P 的切線。AOB 是一條直線，而 APBP 。求

∠POA。

Let OPB = a.

OB = OP (radius)

OBP = OPB (base s, isos. △)

= a

PAB = PBA (base s, isos. △)

= a

OPA = 90 (tangent radius)

In △ABP,

30

180903

180)90(

180

a

a

aaa

BAPBPAABP ( sum of △)

60

3030

OBPOPBPOA (ext. of △)

3. In the figure, TP is the tangent to the circle at P and OQT is a straight

line. If △OPQ is an equilateral triangle, find a.

在圖中，TP 是圓於 P 的切線，而 OQT 是一條直線。若 △OPQ 是

一個等邊三角形，求 a。

POQ = 60 (prop. of equil. △)

OPT = 90 (tangent radius)

In △OPT,

30

1809060

180

a

a

OPTPOTa ( sum of △)

Question Bank

20

4. In the figure, AB is the tangent to the circle at P. If OBOA ,

cm 16AB and the radius of the circle is 6 cm. Find the perimeter of

△OAB.

在圖中，AB 是圓於 P 的切線。若 OBOA ， cm 16AB 及圓的半徑

為 6 cm。求 △OAB 的周界。

Join OP.

∵ OP AB (tangent radius)

∴ OPA = OPB = 90

OP = OP (common side)

OA = OB (given)

∴ △OAP △OBP (RHS)

∴ AP = BP = 8 cm (corr. sides, △s)

In △OPB,

∴ cm 10

86 222

222

OB

OB

OBPBOP

(Pyth. theorem)

∴ The perimeter of △OAB

cm 36

cm )161010(

5. In the figure, , OA // TB and 45OAT . Prove that TB is the tangent to the

circle at T.

在圖中，OA // TB 及 45OAT 。證明 TB 是圓於 T 的切線。

OA = OT radius

OTA = OAT base s, isos. △

= 45

ATB = 45 alt s, TB // OA

90

4545

ATBOTAOTB

∴ TB is the tangent to the circle at T. converse of

tangent radius

6. In the figure, PA is the tangent to the circle at P, AOB is a straight line and

40PAB . Find

(a) ∠POA,

(b) ∠PBA.

在圖中， PA 是圓於 P 的切線，AOB 是一條直線，而 40PAB 。求

(a) ∠POA；


21

(b) ∠PBA。

(a) OPA = 90 (tangent radius)

In △OPA,

50

4090180

180 PAOOPAPOA ( sum of △)

(b)

25

502

1

2

1POAPBA ( at centre twice at ☉ce)

7. In the figure, the diameter of the circle is 30 cm, cm 20TP ,

cm 10AT and TAOB is a straight line. Prove that PT is the tangent to

the circle at P.

在圖中，圓的直徑為 30 cm， cm 20TP ， cm 10AT ，而 TAOB 是一

條直線，證明 PT 是圓於 P 的切線。

Join OP.

cm 15

cm 15

cm 2

30

OP

AO

radius

625

2015 22

22

TPOP

625

25

)1510(

2

22

TO

∴ OP2 + TP2 = TO2

∴ OPT = 90 converse of Pyth. theorem

∴ PT is the tangent to the converse of tangent radius

circle at P.

８. In the figure, AB, BC and CA touches the circle at P, Q and R respectively.

If 62BAC and 40ACB , find PRQ .

在圖中，AB、BC 和 CA 分別與圓相切於 P、Q 和 R。若 62BAC 及

40ACB ，求 PRQ 。

In △RQC,

CR = CQ (tangent properties)

CRQ = CQR (base s, isos. △)

40 + CRQ + CQR = 180 ( sum of △)

Question Bank

22

70

2

40180CRQ

In △APR,

AP = AR (tangent properties)

APR = ARP (base s, isos. △)

62 + APR + ARP = 180 ( sum of △)

59

2

62180ARP

51

7059180

180 CRQARPPRQ (adj. s on st. line)

９. In the figure, TA and TB are tangents to the circle at A and B

respectively. Find x.

在圖中，TA 和 TB 分別是圓於 A 和 B 的切線。求 x。

TA = TB (tangent properties)

x

TABTBA

(base s, isos. △)

In △TAB,

62

2

56180

180562

180

x

x

ATBTABTBA ( sum of △)

１０. In the figure, PA and PB are tangents to the circle at A and B

respectively. If 38APB , find OAB .

在圖中，PA 和 PB 分別是圓於 A 和 B 的切線。若 38APB ，求

OAB 。

PA = PB (tangent properties)

PAB = PBA (base s, isos. △)

In △PAB,

71

2

38180

18038

PAB

PBAPAB ( sum of △)

PAO = 90 (tangent radius)

19

7190

PABPAOOAB

11. In the figure, TP and TQ are tangents to the circle at P and R

respectively. If 40PQT , PSQ and TRQ are straight lines, PS is a

diameter of the circle. Find

(a) PTR ,

(b) RPQ .

在圖中，TP 和 TQ 分別是圓於 P 和 R 的切線。若 40PQT ，PSQ 和 TRQ 都是直線，而 PS


23

則是圓的直徑，求

(a) PTR ；

(b) RPQ 。

(a) QPT = 90 (tangent radius)

In △PQT,

50

4090180

180 PQTQPTPTR

( sum of △)

(b) TP = TR (tangent properties)

TPR = TRP (base s, isos. △)

In △TPR,

PTR + TPR + TRP = 180 ( sum of △)

50 + 2TPR = 180

65

2

50180TPR

25

6590

TPRTPQRPQ

12. In the figure, TA and TB are tangents to the circle at A and B

respectively. Find x.

在圖中，TA 和 TB 分別是圓於 A 和 B 的切線。求 x。

OAT = OBT = 90 (tangent radius)

180

9090

OBTOAT

∴ A, O, B and T are concyclic. (opp. s supp.)

116

ATCAOB (ext. , cyclic quad.)

244

116360

360 AOBx (s at a pt.)

13. In the figure, AP, AB and BR are tangents to the circle at P, Q and R

respectively. If 35RBO and 100AOB , find

(a) PAQ ,

(b) POA .

在圖中，AP、AB 和 BR 分別與圓相切於 P、Q 和 R 。若 35RBO 及

100AOB ，求

(a) PAQ ；

(b) POA 。

(a)

35

RBOQBO (tangent properties)

In △OAB,

Question Bank

24

45

35100180

180 QBOAOBOAQ ( sum of △)

45

OAQPAO (tangent properties)

∴

90

4545PAQ

(b) OPA = 90 (tangent radius)

45

4590180

180 PAOOPAPOA

( sum of △)

14. In the figure, TP and TQ are tangents to the circle at P and Q respectively.

If 58PTQ , TQA and POA are straight lines, find

(a) TPQ ,

(b) QPO ,

(c) PAT .

在圖中，TP 和 TQ 分別是圓於 P 和 Q 的切線。若 58PTQ ，而 TQA

和 POA 都是直線，求

(a) TPQ ；

(b) QPO ；

(c) PAT 。

(a) TP = TQ (tangent properties)

TPQ = TQP (base s, isos. △)

In △TPQ,

58 + TPQ + TQP = 180 ( sum of △)

61

2

58180TPQ

(b) TPA = 90 (tangent properties)

29

6190

TPQTPAQPO

(c) In △TPA,

32

9058180

180 TPAPTQPAT

( sum of △)

15. In the figure, AB is the tangent to the circle at P. If 77RPA ,

35RBP and RQB is straight line, find QPB .

在圖中，AB 是圓於 P 的切線。若 77RPA ， 35RBP ，而

RQB 是一條直線，求 QPB 。

77

RPARQP ( in alt. segment)

In △QPB,


25

42

7735

QPB

QPB

RQPQPBQBP

(ext. of △)

16. In the figure, PQ is the tangent to the circle at T, AOBQ is a straight line.

Find

(a) BTQ ,

(b) BQT .

在圖中，PQ 是圓於 T 的切線，而 AOBQ 是一條直線，求

(a) BTQ ；

(b) BQT 。

(a) ATB = 90 ( in semi-circle)

In △ATB,

30

6090180

180 ABTATBBAT

( sum of △)

30

BATBTQ ( in alt. segment)

(b) BQT + BTQ = ABT (ext. of △)

30

6030

BQT

BQT

17. In the figure, TA and TB are tangents to the circle at A and B respectively.

If 43BAC and 64ATB , find ABC .

在圖中，TA 和 TB 分別是圓於 A 和 B 的切線。若 43BAC 和

64ATB ，求 ABC 。


TAB = TBA (base s, isos. △)

In △TBA,

58

180642

180

TBA

TBA

ATBTBATAB ( sum of △)

58

TBAACB

( in alt. segment)

In △ABC,

79

1805843

180

ABC

ABC

ABCACBBAC ( sum of △)

18. In the figure, AB // PQ and PQ is the tangent to the circle at T, find x.

在圖中，AB // PQ 及 PQ 是圓於 T 的切線，求 x。

. ABT = x (alt. s, AB // PQ)

BAT = x ( in alt. segment)

In △ATB,

Question Bank

26

36

1805

1803

180

x

x

xxx

ATBBATABT

( sum of △)

19. In the figure, AP and AQ are tangents to the circle at P and Q

respectively. If 36PAQ , find QBP .

在圖中， AP 和 AQ 分別是圓於 P 和 Q 的切線。若

36PAQ ，求 QBP 。

AP = AQ (tangent properties)

APQ = AQP (base s, isos. △)

In △APQ,

72

180362

180

AQP

AQP

PAQAQPAPQ ( sum of △)

72

AQPQBP

( in alt. segment)

20. In the figure, a pentagon PQRST is inscribed in a circle and AB is the tangent

to the circle at P. If 38TPA and 110PQR , find RST .

在圖中， PQRST 是一個圓內接於五邊形，而 AB 是圓於 P 的切線。若

38TPA 及 110PQR ，求 RST 。

Join SP.

38

TPATSP ( in alt. segment)

70

180110

180

PSR

PSR

PQRPSR


∴

108

7038

PSRTSPRST

21. In the figure, UTS and QRS are straight lines, 38QPR ,

60PQR and 22USQ . Show that Q, R, T and U are

concyclic.

在圖中，UTS 和 QRS 都是直線。若 38QPR ， 60PQR 及

22USQ ，證明 Q、R、T 和 U 共圓。

In △PQR,


27

82

3860180

180 QPRPQRPRQ

sum of △

In △QSU,

98

2260180

180 USQUQSQUS sum of △

180

9882

QUSPRQ

∴ Q, R, T and U are concyclic. opp. s supp.

22. In the figure, PQ and XY are tangents to the circle at C and A respectively.

If 48PCB and 57XAB , find ADC .

在圖中， PQ 和 XY 分別是圓於 C 和 A 的切線。若 48PCB 及

57XAB ，求 ADC 。

Join BD.

57

XABADB ( in alt. segment)

48

BCPBDC ( in alt. segment)

∴

105

4857

BDCADBADC

23. In the figure, SRPQ and SQPR , show that

(a) △PQR △SRQ,

(b) P, Q, R and S are concyclic.

在圖中， SRPQ 及 SQPR ，證明

(a) △PQR △SRQ；

(b) P、Q、R 和 S 共圓。

(a) PQ = SR given

PR = SQ given

QR = QR common side

△PQR △SRQ SSS

(b) QPR = RSQ corr. s, △s

∴ P, Q, R and S are concyclic. converse of s in

the same segment

Question Bank

28

24. In the figure, 27CAD , 41DCA and 68ABC . Show that A, B, C and D are concyclic

在圖中， 27CAD ， 41DCA 及 68ABC 。證明 A、B、C 和 D 共圓。

In △ACD,

112

4127180

180 DCADACADC sum of △

180

68112

ABCADC

∴ A, B, C and D are concyclic. opp. s supp.

25. In the figure, O is the centre of the circle, AB and CD are chords of the

circle, M and N are mid-points of AB and CD respectively. AB and CD

are produced to meet at P. Show that O, M, P and N are concyclic.

在圖中，O 是圓的圓心。AB 和 CD 是圓上的弦，而 M 和 N 分別是 AB

和 CD 的中點。AB 和 CD 的延線相交於 P。證明 O、M、P 和 N 共

圓。

OMP = 90 line joining centre to

mid-pt. of chord chord

ONP = 90 line joining centre to

mid-pt. of chord chord

∵ OMP + ONP

= 90 + 90

= 180

∴ O, M, P and N are opp. s supp.

concyclic.


29

Level 2 Questions

程度 2 題目

1. In the figure, TA is the tangent to the circle at A and O is the

centre of the circle. M is the mid-point of CD. DCT is a straight

line, TA = 12 cm, TB = 8 cm and OM = 3 cm. Find

(a) OA,

(b) TD.

(Leave your answer in surd form.)

在圖中，TA 是圓於 A 的切線，O 是圓的圓心。M 是 CD 的

中點，DCT 是一條直線，TA = 12 cm，TB = 8 cm 及

OM = 3 cm。求

(a) OA；

(b) TD。

(答案須以根式表示。)

(a) Let OA = r cm.

OB = OA = r cm

OA OT (tangent radius)

5

6416144

)8(12

22

222

222

r

rrr

rr

OTOATA

(Pyth. theorem)

∴ cm 5OA

(b) OM CD (line from centre joining mid-point

of chord chord)

cm 5

OAOB

(radii)

In △TOM,

cm 104

cm 3)58( 22

22

OMOTTM (Pyth. theorem)

Join OD.

In △OMD,

cm 5

OAOD (radii)

cm 4

cm 35 22

22

OMODMD (Pyth. theorem)

Question Bank

30

∴

cm )110(4

cm )4104(

MDTMTD

2. In the figure, O is the centre of the circle. DC is the tangent to the

circle at C and AD cuts the circle at B. If BDC = 60 and

ACD = 74, find AOB.

在圖中，O 為圓的圓心。DC 是該圓於 C 的切線，AD 與圓相交於 B。

若 BDC = 60 及 ACD = 74，求 AOB。

16

7490

90

OCA

OCD

(tangent radius)

OCOA (radii)

16

OCAOAC

(base s, isos. △)

In △ADC,

46

1807460

180

CAD

CAD

CADDCAADC ( sum of △)

OBAO (radii)

OBAOAB (base s, isos. △)

62

4616

CADOACOAB

In △OAB,

56

1806262

180

AOB

AOB

OBAOABAOB

( sum of △)

3. In the figure, PA, PQ and QC are tangents to the circle at A, B and C

respectively. If ABC = 56, find APB + CQB.

在圖中，PA、PQ 和 QC 分別是圓於 A、B 和 C 的切線。若

ABC = 56，求 APB + CQB。

Join AC. Then PA and QC are produced to meet at R.

56

ABCRAC

( in alt. segment)

56

ABCRCA

( in alt. segment)

In △RAC,


31

68

1805656

180

ARC

ARC

RCARACARC

( sum of △)

In △RPQ,

i.e.

180)(68

180)(

180

CQBAPB

CQBAPBARC

RQPRPQPRQ

( sum of △)

112CQBAPB

4. In the figure, PQ and PR are tangents to the circle at A and B

respectively. O is the centre of the circle and DOB is a straight line. If

CAD = 22 and DAQ = 36, find

(a) ADO,

(b) APB.

在圖中，PQ 及 PR 分別是圓於 A 和 B 的切線。O 是圓的圓

心，DOB 是一條直線。若 CAD = 22 及 DAQ = 36，求

(a) ADO；

(b) APB。

(a)

Join OA.

OAQ = 90 (tangent radius)

54

3690

DAQOAQOAD

OD = OA (radii)

54

OADADO (base s, isos. △)

(b) In △OAD,

108

5454

OADADOAOB

(ext. of △)

OAP = OBP = 90 (tangent radius)

72

3601089090

360

APB

APB

AOBOBPOAPAPB

Question Bank

32

5. In the figure, O is the centre of the circle. ODBE is a straight line and

AO OE. EC is the tangent to the circle at C. Prove that ED = EC.

在圖中，O 是圓的圓心。ODBE 是一條直線，且 AO OE。EC

是該圓於 C 的切線。證明 ED = EC。

Join OC.

EDC = ADO vert. opp. s

In △AOD,

90

18090

180

EDCOAC

EDCOAC

ADOAODOAD

sum of △

OA = OC radii

OCAOAC base s, isos. △

∵ 90OCE

90

90

OACECD

OCAECD

tangent radius

∴ ECD + OAC

= OAC + EDC

∴ ECD = EDC

∴ ED = EC sides opp. equal s

6. In the figure, BA and BC are tangents to the circle at A and C

respectively. O is the centre of the circle, OCP and ABP are straight

lines.

(a) Prove that △OAP △BCP.

(b) If AB = 6 cm and BP = 10 cm, find the radius of the circle.

在圖中，BA 和 BC 分別是圓於 A 和 C 的切線。O 是圓的圓心，

OCP 和 ABP 都是直線。

(a) 證明 △OAP △BCP。

(b) 若 AB = 6 cm 及 BP = 10 cm，求圓的半徑。

(a) OAP = 90 tangent radius

PCB = 90 tangent radius

∴ OAD = PCB

CPB = APO common angle

∴ △OAP ~ △BCP AAA

(b)

cm 6

ABBC


In △BPC,


33

cm 8

cm 610 22

22

222

BCBPCP

BPBCCP

(Pyth. theorem)

∵ △OAP ~ △BCP (proved in (a))

BP

OP

CP

AP (corr. sides, ~ △s)

cm 12

cm )820(

cm 102cm 8

2cm 10

cm 8

2cm 10

cm 8

cm 108

106

OC

OC

OC

OC

CPOC

∴ The radius of the circle is 12 cm.

7. In the figure, the circle is inscribed in △ABC and touches the sides of the

triangle at P, Q and R, AQ = 10 cm, QC = 3 cm and ABC = 90. Find BR.

在圖中，一個圓內接於 △ABC，且與三角形相切於 P、Q 和 R，

AQ = 10 cm，QC = 3 cm 及 ABC = 90。求 BR。

. Let BR = x cm.

cm x

BPBR


cm 10

AQAR


cm 3

QCPC


In △ABC,

BC2 + AB2 = AC2 (Pyth. theorem)

0)15)(2(

03013

060262

13)10()3(

)()()(

2

2

222

222

xx

xx

xx

xx

QCAQRBARPCBP

x = 2 or x = 15 (rejected)

∴ cm 2BR

8. In the figure, ABCD is a cyclic quadrilateral. PQ and PR are tangents

to the circle at A and C respectively. If PAB = 36 and PCB = 28,

find

(a) ADC,

(b) APC.

在圖中，ABCD 是一圓內接四邊形。PQ 和 PR 分別是圓於 A 和

C 的切線。若 PAB = 36 及 PCB = 28，求

(a) ADC；

(b) APC。

Question Bank

34

(a)

Join BD.

28

BCPBDC

( in alt. segment)

36

BAPADB

( in alt. segment)

64

3628

ADBBDCADC

(b)

Join AC.

64

ADCPAC

( in alt. segment)

64

ADCPCA

( in alt. segment)

In △PAC,

180PCAPACAPC ( sum of △)

52

128180

1806464

APC

APC

9. In the figure, PA and PC are tangents to the circle at A and C

respectively. O is the centre of the circle, PQC and ABQ are

straight lines, OBC = 62 and AQ PC. Find

(a) BAC,

(b) APQ.

在圖中，PA 和 PC 分別是圓於 A 和 C 的切線。O 是圓的圓

心，PQC 和 ABQ 都是直線，OBC = 62 及 AQ PC。求

(a) BAC；

(b) APQ。

(a) OB = OC (radii)

62

OBCOCB

(base s, isos. △)

In △BOC,


35

56

6262180

180 OBCOCBBOC

( sum of △)

28

256

2

BAC

BAC

BACBOC

( at centre twice at ⊙ce)

(b) In △AQC,

62

1802890

180

ACQ

ACQ

QACCQAACQ

( sum of △)

∵ PC = PA (tangent properties)

∴

62

ACQPAC (base s, isos. △)

In △APC,

56

1806262

180

APQ

APQ

ACQPACAPQ

( sum of △)

10. In the figure, TC is the tangent to the circle at C. O is the centre of

the circle. DB and DC bisect OBC and OCB respectively and

BCT = 67. Find

(a) BOC,

(b) BDC.

在圖中，TC 是圓於 C 的切線。O 是圓的圓心。DB 和 DC 分

別平分 OBC 和 OCB，且 BCT = 67。求

(a) BOC；

(b) BDC。

(a)

67

BCTBAC

( in alt. segment)

134

672

2 BACBOC


(b) In △OBC,

46

180134

180

OCBOBC

OCBOBC

OCBOBCBOC

( sum of △)

∵ DB and DC bisect OBC and OCB respectively.

∴

23

462

1

)(2

1OCBOBCDCBDBC

In △BDC,

157

18023

180

BDC

BDC

DCBDBCBDC

( sum of △)

Question Bank

36

11. In the figure, TS is the tangent to the circle at A, BA = BC,

BAT = 36 and 5:3:

DCAD . Find

(a) ABC,

(b) SAD.

在圖中，TS 是圓於 A 的切線，BA = BC，BAT = 36 及

5:3:

DCAD 。求

(a) ABC；

(b) SAD。

(a)

36

BATACB ( in alt. segment)

∵ BA = BC

∴

36

ACBCAB

(base s, isos. △)

In △ABC,

108

3636180

180 CABACBABC

( sum of △)

(b)

72

180108

180

CDA

CDA

CDAABC


DAC

DCA

DAC

DCA

DC

AD

5

2

(arcs prop. to s at ⊙ce)

∴ DCA = 3k and DAC = 5k, where k 0

In △ACD,

5.13

1088

1803572

180

k

k

kk

DCADACCDA ( sum of △)

∴

5.40

5.133PCA

5.40

PCASAD ( in alt. segment)

12. In the figure, DB and DF are tangents to the circle at B and C

respectively. BD and AC are produced to meet at E. If

ABC = 52 and CED = 34, find

(a) DCE,

(b) CAB.

在圖中，DB 和 DF 分別是圓於 B 和 C 的切線。BD 和 AC

的延線相交於 E。若 ABC = 52 及 CED = 34，求

(a) DCE；

(b) CAB。


37

(a)

52

ABCACF

( in alt. segment)

52

ACFDCE (vert. opp. s)

(b) In △CDE,

86

3452

CEDDCECDB

(ext. of △)

In △BDC,

BD = DC (tangent properties)

DBC = DCB (base s, isos.

△)

47

180862

180

DBC

DBC

CDBDCBDBC ( sum of △)

47

DBCCAB

( in alt. segment)

13. In the figure, TA is the tangent to the circle at A, TBC is a straight line,

TB = 10 cm and BC = 30 cm.

(a) Prove that △TAB △TCA.

(b) Hence, find TA.

在圖中，TA 是圓於 A 的切線，TBC 是一條直線，TB = 10 cm 及

BC = 30 cm。

(a) 證明 △TAB △TCA。

(b) 由此，求 TA。

(a) ATB = CTA common angle

TAB = TCA in alt. segment

∴ △TAB ~ △TCA AAA

(b) ∵ △TAB ~ △TCA

∴

cm 20

cm 400

10

103022

TA

TA

TA

TA

TA

TB

TC

TA(corr.

14. In the figure, AB is a diameter of the circle. PQ is the tangent to the

circle at C, AC and BD intersect at K and BKC = 48.

(a) Find DCP.

(b) If 3:2:

DCAD , find BCQ.

在圖中，AB 是圓的直徑。PQ 是圓於 C 的切線，AC 和 BD 相

交於 K，BKC = 48。

(a) 求 DCP。

(b) 若 3:2:

DCAD ，求 BCQ。

Question Bank

38

(a) ABC = 90 ( in semi-circle)

In △KBC,

42

1809048

180

CBK

CBK

KCBCBKCKB

( sum of △)

42

CBKDCP

( in alt. segment)

(b)

CBK

ABD

DBC

ABD

DC

AD

3

2


∵

42

DCPDBC

( in alt. segment)

∴

28

3

242

423

2

ABD

ABD

In △ABC,

20

18090)4228(

180

CAB

CAB

ACBABCCAB

( sum of △)

20

CABBCQ

( in alt. segment)

15. In the figure, TA and TC are tangents to the circle at A and C respectively. If

5:6:7::

CABCAB , find

(a) ABC,

(b) ATC.

在圖中，TA 和 TC 分別是圓於 A 和 C 的切線。若

5:6:7::

CABCAB ，求

(a) ABC；

(b) ATC。

(a) ∵

︵AB :

︵BC :

︵CA = 7 : 6 : 5

∴ BCA : BAC : ABC = 7 : 6 : 5


∴ BCA = 7k, BAC = 6k, ABC = 5k, where k 0

In △ABC,

10

18018

180765

180

k

k

kkk

BCABACABC ( sum of △)

∴

50

105ABC

(b)

50

ABCACT ( in alt. segment)


39

50

ABCTAC ( in alt. segment)

In △ACT,

80

1805050

180

ATC

ATC

TACACTATC

( sum of △)

16. In the figure, O is the centre of the circle. DE is the tangent to the circle

at D. BE cuts the circle at C and AOB is a straight line. If OBC = 64

and BE DE, find CDE.

在圖中，O 是圓的圓心。 DE 是圓於 D 的切線。BE 與圓相交於

C，且 AOB 是一條直線。若 OBC = 64 及 BE DE，求 CDE。

.

Join AC and let CDE = x.

x

CDEDAC

( in alt. segment)

116

18064

180

ADC

ADC

ABCADC


ACB = 90 ( in semi-circle)

CED = 90 (given)

∴ AC // DE (corr. s equal)

x

CDEACD

(alt. s, AC // DE)

In △ACD,

32

642

1801162

180116

180

x

x

x

xx

ACDDACADC

( sum of △)

∴ 32CDE

17. In the figure, TA is the tangent to the circle at A, TBD is a straight

line and CA is diameter of the circle. If ATB = 38 and

CDB = 54, find BCD.

. 在圖中，TA 是圓於 A 的切線，TBD 是一條直線，CA 是該

圓的一條直徑。若 ATB = 38 及 CDB = 54，求 BCD。

Question Bank

40

Join AB.

54

CDBCAB (s in the same segment)

90CAT (tangent radius)

36

5490

CABCATBAT

36

BATBCA

( in alt. segment)

In △ABT,

74

3638

BATBTAABD

(ext. of △)

74

ABDACD


110

7436

ACDBCABCD

18. In the figure, TP and TR are tangents to the circle at P and R

respectively. RSA is a straight line, PQ // AR, QPR = 38 and

PTR = 62. Find

(a) PAR,

(b) PSR.

在圖中，TP 和 TR 分別是圓於 P 和 R 的切線。RSA 是一條

直線，PQ // AR，QPR = 38 及 PTR = 62。求

(a) PAR；

(b) PSR。

(a)

38

QPRPRS

(alt. s, AR // PQ)

38

PRSSPA

( in alt. segment)

ART = RPS ( in alt. segment)

In △PRT,

180TPRRTPPRT ( sum of △)

21

180386238

180)(62)(

ART

ARTART

RPSTPSARTPRA

In △ART,

83

2162

ARTATRPAR

(ext. of △)


41

(b) In △PSA,

121

8338

PARSPAPSR

(ext. of △)

19. In the figure, O is the centre of the circle. TS is the tangent to the circle at C.

AOBT and ADS are straight lines, AS TS.

(a) Prove that BAC = CAS.

(b) If BTC = 44, find ADC.

在圖中，O 是圓的圓心。TS 是圓於 C 的切線。AOBT 和 ADS 都是

直線，AS TS。

(a) 證明 BAC = CAS。

(b) 若 BTC = 44，求 ADC。

. (a)

CAS

BCA

90

( in semi-circle)

ACS = ABC ( in alt. segment)

∴ △BAC ~ △CAS (AAA)

∴ BAC = CAS (corr. s, ~ △s)

(b) In △ATS,

23

462

180)(9044

180

CAB

CAB

CABSAC

SATTSABTC ( sum of △)

In △ABC,

67

1809023

180

CBA

CBA

CBAACBCAB

( sum of △)

113

18067

180

ADC

ADC

CBAADC


20. In the figure, PQ is the tangent to the circle at C. DB is a

diameter of the circle. BD and AC intersect at K, AB // PQ,

BKC = 81 and DCK = x. Find x.

在圖中，PQ 是圓於 C 的切線。DB 是圓的一條直徑。

BD 和 AC 相交於 K，AB // PQ，BKC = 81 及

DCK = x。求 x。

DCP = DBC ( in alt. segment)

DBCx

ACPBAC

(alt. s, AB // PQ)

x

ACD

ABDABK


In △ABK,

xDBC

xDBCx

BKCABKBAK

281

81)(

(ext. of △)

(proved in (a))

Question Bank

42

x

xx

DBCx

BACBDC

81

)281(


90DCB ( in semi-circle)

In △BDC,

24

372

180)281(90)81(

180

x

x

xx

CBDDCBBDC

( sum of △)

Alternative Solution

BCD = 90 ( in semi-circle)

In △DCK,

xCDK

BKCDCKCDK

81

(ext. of △)

In △BDC,

9

180)81(90

180

xDBC

xDBC

BDCBCDDBC

( sum of △)

x

CDK

CDBBAC

81


9x

DBCDCP ( in alt. segment)

∵ ACP = BAC (alt. s, PQ // AB)

∴

24

723

81)9(

x

x

xxx

BACxDCP

21. In the figure, PQ and RS are tangents to the circle at A and B respectively,

AP = AC and PC // RS.

(a) Prove that PA = PK.

(b) If APK = 34, find ABC.

在圖中，PQ 和 RS 分別是圓於 A 和 B 的切線，AP = AC 及

PC // RS。

(a) 證明 PA = PK。

(b) 若 APK = 34，求 ABC。

(a) PAB = ACB ( in alt. segment)

RBK = ACB ( in alt. segment)

PKA = RBK (corr. s, PC // RS)

∴ PAB = PKA

∴ PA = PK (sides opp. equal s)

(b) AC = AP

34

APK

APCACP

(base s, isos. △)

In △ACP,

68

3434

APCACPCAQ

(ext. of △)


43

68

CAQABC

( in alt. segment)

22. In the figure, BF is the tangent to the circle at B.

ADC = 115, DFE = 40,

CDBC , ADF and BEF

are straight lines.

(a) Find CBE.

(b) Prove that C, D, F and E are concyclic.

在圖中，BF 是圓於 B 的切線。ADC = 115，

DFE = 40，

CDBC ，ADF 和 BEF 都是直線。

(a) 求 CBE。

(b) 證明 C、D、F 和 E 共圓。

(a) Let CBE = a.

65

180115

180

ABC

ABC

ADCABC


a

CBECAB

( in alt. segment)

∵

︵BC =

︵CD

∴

a

CABDAC


In △ABF,

25

18040)65(2

180

a

aa

AFBABFBAD

( sum of △)

∴ 25CBE

(b)In △ACD,

140

25115

DACCDADCE

ext. of △

180

40140

DFEDCE

∴ C, D, F and E are concyclic. opp. s supp.

23. In the figure, PQ and PR are tangents to the circle at A

and C respectively. O is the centre of the circle,

OAB = 19 and AB = AC. Find

(a) ABC,

(b) APC.

在圖中，PQ 和 PR 分別是圓於 A 和 C 的切線。O

是圓的圓心，OAB = 19 及 AB = AC。求

(a) ABC；

(b) APC。

Question Bank

44

(a)

Join OB.

OA = OB (radii)

19

OABOBA

(base s, isos. △)

In △AOB,

142

1801919

180

AOB

AOB

OBAOABAOB

( sum of △)

∵ AB = AC

AOC = AOB = 142 (equal s, equal chords)

76

360142142

360

COB

COB

AOBAOCCOB

(s at a pt.)

OC = OB (radii)

OCB = OBC (base s, isos. △)

In △OCB,

52

180762

180

OBC

OBC

COBOCBOBC

( sum of △)

71

5219

OBCABOABC

(b)

71

ABCPAC

( in alt. segment)

71

ABCPCA ( in alt. segment)

In △APC,

38

1807171

180

APC

APC

PCAPACAPC

( sum of △)

24. In the figure, CB is a diameter of the circle and PQ is the tangent to the

circle at A, CDB is a straight line and BAQ = BAD = x.

(a) Express ABC in terms of x.

(b) Prove that

(i) AD BC,

(ii) CA bisects PAD.

在圖中，CB 是圓的一條直徑，PQ 是圓於 A 的切線，CDB 是一

條直線，BAQ = BAD = x。

(a) 以 x 表示 ABC。

(b) 證明


45

(i) AD BC；

(ii) CA 平分 PAD。

(a) CAB = 90 ( in semi-circle)

x

BAQACB

( in alt. segment)

In △ABC,

xABC

xABC

BCABACABC

90

18090

180

( sum of △)

(b) (i) Consider △ADB and △CAB.

BAD = BCA = x proved in (a)

ABD = ABC common angle

∴ △ADB ~ △CAB AAA

∴ ADB = CAB = 90 corr. s, ~ △s

∴ AD BC

(ii) In △ACD,

xCAD

CADx

CADADCACD

90

18090

180

sum of △

PAC = ABD in alt. segment

In △ABD,

xPAC

PACx

ABDADBDAB

90

18090

180

sum of △

∴ CAD = PAC

∴ CA bisects PAD.

25. In the figure, PQ is the tangent to the circle at A. BD is a diameter

of the circle. If BAP = 55 and BRC = 18, find

(a) BAC,

(b) BPA.

在圖中，PQ 是圓於 A 的切線。BD 是該圓的一條直徑。若

BAP = 55 及 BRC = 18，求

(a) BAC；

(b) BPA。

. (a)

BAC

BACCRADCA

18

(ext. of △)

BAC

DCADAQ

18

( in alt. segment)

DAB = 90 ( in semi-circle)

17

1805590)18(

180

BAC

BAC

BAPDABDAQ (adj. s on st. line)

(b) BCA = 90 ( in semi-circle)

In △CBR,

72

9018

CBR

CBR

BCDBRCCBR

(ext. of △)

Question Bank

46

72

CBRABP

(vert. opp. s)

In △ABP,

53

1805572

180

BPA

BPA

BAPABPBPA

( sum of △)

26. In the figure, PQ is the tangent to the circle at A. DC // PQ, BA = BC

and DAP = 74. Find

(a) ABC,

(b) BAD.

在圖中，PQ 是圓於 A 的切線。DC // PQ，BA = BC 及

DAP = 74。求

(a) ABC；

(b) BAD。

(a)

74

DAPADC

(alt. s, DC // PQ)

106

18074

180

ABC

ABC

ABCADC (opp. s, cyclic quad.)

(b)

Join AC.

∵ BA = BC

∴ BAC = BCA (base s, isos. △)

In △ABC,

37

742

1802106

180

BCA

BCA

BCA

BACBCAABC

( sum of △)

37

BCABAQ

( in alt. segment)

69

1803774

180

BAD

BAD

BAQBADDAP (adj. s on st. line)

27. In the figure, ABCD is a quadrilateral and its diagonals intersect at K.

KAD = 20, KAB = 40 and ABC : BCD : CDA = 5 : 6 : 4.

(a) Prove that ABCD is a cyclic quadrilateral.

(b) Find BKC.


47

在圖中，四邊形 ABCD 的對角線相交於 K，KAD = 20，

KAB = 40 及 ABC : BCD : CDA = 5 : 6 : 4。

(a) 證明 ABCD 是一個圓內接四邊形。

(b) 求 BKC。

(a) ∵ ABC : BCD : CDA = 5 : 6 : 4

∴ ABC = 5k, BCD = 6k

and CDA = 4k, where k 0

20

30015

360)4020(465

360

k

k

kkk

DABCDABCDABC

∴ ABC = 100, BCD = 120 and

CDA = 80

∵ CDA + CBA

= 80 + 100

= 180

∴ ABCD is a cyclic quadrilateral. opp. s supp.

(b) ∵ ABCD is a cyclic quadrilateral.

∴

20

CADCBD (s in the same segment)

80

20100

CBDABCABD

In △ABK,

120

4080

BAKKBABKC

(ext.

28. In the figure, ABCD is a quadrilateral and its diagonals intersect at K.

ADK = 37, AKB = 75, DCK = 33 and BCD = 70.

(a) Prove that ABCD is a cyclic quadrilateral.

(b) Find ABC.

. 在圖中，四邊形 ABCD 的對角線相交於 K。若 ADK = 37，

AKB = 75，DCK = 33 及 BCD = 70，

(a) 證明 ABCD 是一個圓內接四邊形；

(b) 求 ABC。

(a)

37

3370

DCADCBACB

∴ ADB + ACB

∴ ABCD is a cyclic converse of s in

quadrilateral. the same segment

(b) ∵ ABCD is a cyclic quadrilateral.

∴

33

ACDABD (s in the same segment)

In △BKC,

Question Bank

48

38

3775

ACBAKBKBC

(ext. of △)

71

3833

KBCABDABC

29. In the figure, O is the centre of the circle, CO AB, OP = OQ, APOB and

CQO are straight lines.

(a) Prove that △COP △BOQ.

(b) Hence, or otherwise, prove that O, B, C and R are concyclic.

在圖中，O 是圓的圓心，CO AB，OP = OQ，APOB 和 CQO 都是

直線。

(a) 證明 △COP △BOQ。

(b) 由此，或用其他方法，證明 O、B、C 和 R 共圓。

(a) OB = OC radii

QOB = COP = 90 given

OP = OQ given

∴ △COP △BOQ SAS

(b) ∵ △COP △BOQ proved in (a)

∴ OCP = OBQ corr. s, △s

∴ O, B, C and R are converse of s in

concyclic. the same segment

30. In the figure, PQ is the tangent to the circle at A. O is the centre of the circle,

OQ intersect AC at K.

(a) Prove that ABOK is a cyclic quadrilateral.

(b) Hence, or otherwise, prove that AQ = KQ.

在圖中，PQ 是圓於 A 的切線。O 是圓的圓心，OQ 與 AC 相交於 K。

(a) 證明 ABOK 是一個圓內接四邊形。

(b) 由此，或用其他方法，證明 AQ = KQ。

(a) BAC = 90 in semi-circle

∵ BAC = KOC = 90

∴ ABOK is a cyclic ext. = int. opp.

quadrilateral.

(b) ∵ ABOK is a cyclic

quadrilateral. proved in (a)

∴ AKQ = ABO in alt. segment

ABO = KAQ in alt. segment

KAQ = AKQ

∴ AQ = KQ sides opp. equal s


49

31. In the figure, O is the centre of the circle. ADC is a straight line and

DA = DB.

(a) Prove that BDC = 2BAD.

(b) Show that O, D, B and C are concyclic.

在圖中，O 是圓的圓心。ADC 是一條直線及 DA = DB。

(a) 證明 BDC = 2BAD。

(b) 證明 O、D、B 和 C 共圓。

(a) ∵ AD = DB

∴ DAB = DBA base s, isos. △

In △ABD,

BADBDC

DBADABBDC

2

ext. of △

(b)

Join OB and OC.

BACBOC 2 at centre twice at ⊙ce

BDC

BAD

2

proved in (a)

∴ O, D, B and C converse of s in

are concyclic. the same segment

32. In the figure, the two circles intersect each other at C and D. PAB, PEF,

ACF and BCE are straight lines. Prove that PADF is a cyclic quadrilateral.

在圖中，兩個圓相交於 C 和 D。PAB、PEF、ACF 和 BCE 都是直

線。證明 PADF 是一個圓內接四邊形。

ABC = ADC s in the same segment

PEC = CDF ext. , cyclic quad.

In △PBE,

PBE + PEB + BPE = 180 sum of △

∴

180

180

APEADF

APECDFABC

∴ PADF is a cyclic quadrilateral. opp. s supp.

33. In the figure, RQ is the tangent to the circle at Q and O is the

centre of the circle. SBOQ is a straight line and BA // SR.

(a) Prove that PQRS is a cyclic quadrilateral.

(b) Hence, or otherwise, find SPR.

在圖中，RQ 是圓於 Q 的切線，O 是該圓的圓心。SBOQ 是

一條直線，且 BA // SR。

(a) 證明 PQRS 是一個圓內接四邊形。

(b) 由此，或用其他方法，求 SPR。

Question Bank

50

(a) PQB = PAB s in the same segment

PAB = PRS corr. s, BA // SR

∴ PQB = PRS

∴ PQRS is a cyclic converse of s in the

quadrilateral. same segment

(b) SQR = 90 (tangent radius)

90

SQRSPR


34. In the figure, P and Q are the centres of the circles ABD and ABC

respectively. PBC is a straight line. Denote ABP by x.

(a) Express ADB and AQC in terms of x.

(b) Are A, P, C and Q concyclic? Give the reason.

. 在圖中，P 和 Q 分別是圓 ABD 和圓 ABC 的圓心。PBC 是一

條直線。設 ABP 為 x。

(a) 以 x 表示 ADB 和 AQC。

(b) A、P、C 和 Q 是否共圓？試說明理由。

. (a) PA = PB (radii)

x

PBAPAB

(base s, isos. △)

In △APB,

xAPB

xxAPB

PBAPABAPB

2180

180

180

( sum of △)

Mark E on the circle ABC.

Join AE and EC.

x

ABPAEC

(ext. , cyclic quad.)

x

APCAQC

2

2


(b) APB + AQC

= 180 – 2x + 2x

= 180

∴ A, P, C and Q are concyclic. opp. s supp.


51

Level 2+ Questions

程度 2+ 題目

1. In the figure, PA and PB are tangents to the circle AQB at A

and B respectively. PC and PD are tangents to the circle CRD

at C and D respectively. AQSTRD, PSB and PTC are straight

lines. It is given that PBC = PCB = y, APB = x and

CPD = z.

(a) Prove that PA = PD.

(b) If BC // AD, prove that x = z.

. 在圖中，PA 和 PB 分別是圓 AQB 於 A 和 B 的切線。而 PC 和 PD 則分別是圓 CRD 於 C

和 D 的切線。AQSTRD、PSB 和 PTC 都是直線。已知

PBC = PCB = y，APB = x 及 CPD = z。

(a) 證明 PA = PD。

(b) 若 BC // AD，證明 x = z。

(a) PD = PC tangent properties

PA = PB tangent properties

PBC = PCB given

PB = PC sides opp. equal s

∴ PA = PD

(b)

y

PBCPST

corr. s, AD // BC

y

PCBPTS

corr. s, AD // BC

PA = PD proved in (a)

PAD = PDA base s, isos. △

In △PAS,

PAS + APS = PST ext. of △

∴ PAS + x = y

∴ PAD + x = y

In △PDR,

TPD + PDT = PTS ext. of △

∴ z + PDT = y

∴ z + PDA = y

∴ PAD + x = PDA + z

∵ PAD = PDA

∴ x = z

2. In the figure, PQ is the tangent to the circle at D. O is the centre of

the circle, PAOB and BCQ are straight lines, BQ PQ and

OAD = 56.

(a) Find CDQ.

(b) Is CD = DA? Give the reason.

(c) Is CD // PB? Give the reason.

在圖中，PQ 是圓於 D 的切線。O 是該圓的圓心，PAOB 和 BCQ 都是直線，而 BQ PQ 及

Question Bank

52

OAD = 56。

(a) 求 CDQ。

(b) CD 是否等於 DA？試說明理由。

(c) CD 是否平行於 PB？試說明理由。

(a) DABDCQ (ext. , cyclic quad.)

86 In △CDQ,

180DCQCQDCDQ ( sum of △)

34

1805690CDQ

(b)

Join AC.

90ACB ( in semi-circle)

124

18056

180

DCB

DCB

BADDCB


34

90124

ACBDCBACD

34

CDQDAC

( in alt. segment)

∴ DACACD

∴ DACD (sides opp. equal s)

(c) ∵ 34DCADAC

22

3456

CADBADBAC

∵ ACDBAC

∴ CD is not parallel to PB.

3. In the figure, PA is the tangent to the circle at A and AC // PQ.

CBQ and PQR are straight lines.

(a) Prove that △RBQ △RPA.

(b) Prove that △ABC △RPA.

(c) If RB = 2 cm, BA = 4 cm, CA = 3 cm and AP = 5 cm, find

BQ.

. 在圖中，PA 是圓於 A 的切線，而 AC // PQ。CBQ 和 PQR

都是直線。

(a) 證明 △RBQ △RPA。

(b) 證明 △ABC △RPA。


53

(c) 若 RB = 2 cm，BA = 4 cm，CA = 3 cm 及 AP = 5 cm，求 BQ。

∵ △ABC ~ △RPA (proved in (b))

cm 5.2

cm 52

1

cm6

3

cm 5

cm24

3

cm 5

CB

CB

CB

RA

AC

AP

CB

(corr. sides, ~△s)

By (a) and (b), △RBQ ~△ABC

cm 25.1

cm 5.22

1

cm4

2

cm 5.2

BQ

BQ

BA

BR

BC

BQ

(corr. sides, ~△s)

4. In the figure, the circle touches the quadrilateral PQRS at the points

A, B, C and D. MPBQ and NSDR are straight lines, MPA = 80,

NSA = 116 and BA = BC.

(a) Find BAD.

(b) Find ADC.

(c) Is PQRS a cyclic quadrilateral? Give the reason.

在圖中，四邊形 PQRS 與圓相切於 A、B、C 和 D。 MPBQ 和 NSDR 都是直線，MPA = 80，

NSA = 116 及 BA = BC。

(a) 求 BAD。

(b) 求 ADC。

(c) PQRS 是否圓內接四邊形？試說明理由。

(a) SDAS (tangent properties)

SDASAD (base s, isos. △)

In △ASD,

58

1162

SAD

SAD

ASNSDASAD

(ext. of △)

PBPA (tangent properties)

PBAPAB

In △PAB,

40

802

PAB

PAB

APMPBAPAB

(ext. of △)

82

4058180

180 PABSADBAD

(adj. s on st. line)

Question Bank

54

(b)

Join AC.

40PBA (from (a))

BAC = 40 ( in alt. segment)

BABC

40

BACBCA

(base s, isos. △)

In △ABC,

100

4040180

180 BACBCAABC

180ABCADC (opp. s, cyclic quad.)

80

180100

ADC

ADC

(c) 58ADS (from (a))

42

1808058

180

CDR

CDR

CDRADCADS (adj. s on st. line)

RDRC (tangent properties)

42

RDCRCD

(base s, isos. △)

In △RCD,

96

1804242

180

DRC

DRC

DRCRDCRCD ( sum of △)

∵ 80MPS and 96DRC

∴ DRCMPS

∴ PQRS is not a cyclic quadrilateral.

5. In the figure, O is the centre of the circle BDC. ABOC is a

straight line and AD is the tangent to the circle at D.

(a) Prove that △ABD △ADC.

(b) If BD = 6 cm, CD = 8 cm, AB = x cm and AD = y cm,

(i) prove that 104

3

x

y

y

x,

(ii) find x and y.

在圖中，O 是圓 BDC 的圓心。ABOC 是一條直線，而 AD 則是該圓於 D 的切線。

(a) 證明 △ABD △ADC。

(b) 若 BD = 6 cm，CD = 8 cm，AB = x cm 及 AD = y cm，

(i) 證明 104

3

x

y

y

x，


55

(ii) 求 x 和 y。

(a)

BDA = DCA in alt. segment

BAD = DAC common angle

∴ △ABD ~ △ADC AAA

(b) (i) ∵ △ABD ~ △ADC

∴

y

x

y

x

AD

AB

DC

BD

4

3

8

6

corr. sides, ~△s

In △BDC,

BDC = 90 in semi-circle

cm 10

cm 86 22

22

DCBDBC

Pyth. theorem

104

3

108

6

x

y

x

y

AC

AD

DC

BD

corr. sides, ~△s

∴ 104

3

x

y

y

x

(ii) From (b)(i),

y

xy

y

x

4

4

3

By substituting 3

4xy into

104

3

x

y, we have

∴

7

90

907

16909

3

16303

10

3

4

4

3

x

x

xx

xx

x

x

∴

7

120

7

90

3

4

y

6. In the figure, ABCD is a straight line and C is the centre of the

circle. AF and FD are tangents to the circle at E and D respectively.

(a) Prove that AEB = AFC.

(b) Prove that △ABE △ACF.

(c) If AB = 4 and BD = 12, find AE and EF.

Question Bank

56

在圖中，ABCD 是一條直線，而 C 是圓的圓心。AF 和 FD 分別是該圓於 E 和 D 的切線。

(a) 證明 AEB = AFC。

(b) 證明 △ABE △ACF。

(c) 若 AB = 4 及 BD = 12，求 AE 和 EF。

(a) Let EFC = x.

x

EFCCFD

tangent properties

CDF = 90 tangent radius

In △AFD,

180DAFADFAFD sum of △

xDAF

DAFx

DAFCFDEFC

290

180902

18090)(

CEF = 90 tangent radius

∵

180

9090CDFCEF

∴ C, D, E and F are concyclic. opp. s supp.

∴

x

EFDACE

2

ext. , cyclic quad.

CB = CE radii

CBE = CEB base s, isos. △

In △CBE,

xCBE

CBEx

CEBCBEBCE

90

18022

180

sum of △

In △ABE,

AFC

xAEB

xAEBx

CBEAEBEAB

90)290(

ext. of △

(b) EAB = FAC common angle

AEB = AFC proved in (a)

∴ △ABE ~ △ACF AAA

(c) ∵ BD is a diameter of the circle.

∴

cm 6

cm 2

12

2

BD

BC

∵ △ABE ~ △ACF

∴

5

2

64

4

BCAB

AB

EFAE

AE

AC

AB

AF

AE (corr. sides, ~△s)

EFAE

EFAEAE

EFAEAE

23

225

)(25

EFAE3

2 ……(1)

FD = EF (tangent properties)


57

In △AFD,

AF2 = FD2 + AD2 (Pyth. theorem)

222 16)( EFEFAE ……(2)

By substituting (1) into (2), we have

12

144

169

16

169

25

163

5

163

2

2

22

222

22

2

22

2

EF

EF

EF

EFEF

EFEF

EFEFEF

8

123

2

AE

∴ AE = 8 and EF = 12

7. the figure, the circle ABCD intersects the circle ADEF at A and

D. PE is the tangent to the circle ABCD at A. BDE and CDF are

straight lines.

(a) Prove that AC // FE.

(b) Prove that BAC = FAE.

(c) If AFE = 90, prove that BAE = 90.

在圖中，ABCD 與圓 ADEF 相交於 A 和 D。PE 是圓

ABCD 於 A 的切線。BDE 和 CDF 都是直線。

(a) 證明 AC // FE。

(b) 證明 BAC = FAE。

(c) 若 AFE = 90，證明 BAE = 90。

(a) ABDEAD in alt. segment

ACDABD sin the same segment

∴ ACDEAD

CFEEAD sin the same segment

∴ CFEACD

∴ AC // FE alt. s equal

(b) BDCBAC sin the same segment

FDEFAE sin the same segment

FDEBDC vert. opp. s

∴ FAEBAC

(c) 180AFECAF int. s, AC // FE

90

18090

CAF

CAF

given

FAEBAC proved in (b)

∴

90

CAF

CAEFAE

CAEBACBAE

Question Bank

58

8. In the figure, PA is the tangent to the circle at A. OBP and DMCP

are straight lines. PA = 24 cm, PB = 16 cm, PC = 18 cm, OB = r cm

and DM = MC = x cm.

(a) Find r.

(b) By considering △OMD and △OMP, find the value of x.

(c) Is DOA a straight line? Give the reason.

在圖中，PA 是圓於 A 的切線。OBP 和 DMCP 都是直線。

PA = 24 cm，PB = 16 cm，PC = 18 cm，OB = r cm 及

DM = MC = x cm。

(a) 求 r。

(b) 利用 △OMD 和 △OMP，求 x 的值。

(c) DOA 是否直線？試說明理由。

. (a) 90OAP (tangent radius)

cm rOA (radius)

In △OAP,

OA2 + AP2 = OP2 (Pyth. theorem)

r2 + 242 = (r + 16)2

r2 + 576 = r2 + 32r + 256

32r = 320

r =10

(b) ∵ DM = MC = x cm

∴ OMD = 90 (line joining centre and mid-point

of chord bisects chord)

In △ODM,

OM2 = OD2 DM2 (Pyth. theorem)

= r2 x2

= 100 x2

In △OMP,

OM2 = OP2 MP2 (Pyth. theorem)

= (r + 16)2 (x + 18)2

= 262 (x + 18)2

= 676 x2 36x 324

= 352 x2 36x

∴

7

25236

36352100 22

x

x

xxx

(c)

2

22

2222

cm 976

cm ]24)1010[(

)(

APOADOAPDA

2

222

cm 1024

cm)1872(

DP

∴ 222 DPAPDA

∴ DOA is not a straight line.


59

Multiple Choice Questions

多項選擇題

1. In the figure, which of the following must be

true?

I. A, R, P and Q are concyclic.

II. B, P, Q and R are concyclic.

III. C, Q, R and B are concyclic.

A. I only

B. II only

C. III only

D. none of them

在圖中，下列何者

必為正確？

I. A、R,、P 和 Q 共圓。

II. B、P、Q 和 R 共圓。

III. C、Q、R 和 B 共圓。

A. 只有 I

B. 只有 II

C. 只有 III

D. 以上選擇皆不正確

A

In △PCQ,

90

4050180

180 QCPQPCPQC sum of △

90

90180

180 BRPARP adj. s on st. line

∴ A, R, P and Q are concyclic. ext. = int. opp.

∴ I is true.

In quadrilateral BPQR,

50

110

9020

QPC

QRB

∴ QRB QPC

∴ II is not true.

In quadrilateral CQRB,

180

150

40)2090(

PCQBRQ

∴ III is not true.

2 In the figure, BC is the tangent to the circle at

C and O is the centre of the circle. If

AB = 4 cm and BC = 8 cm, find OC.

在圖中，BC 是圓於 C 的切線，而 O 則

是該圓的圓心。若 AB = 4 cm 及

BC = 8 cm，求 OC。

A. 5 cm

B. 5.5 cm

C. 6 cm

D. 6.5 cm

C Let the radius of the circle is r cm.

OA = OC (radii)

= r cm

6

64816

8)4(

22

222

222

r

rrr

rr

COBCOB (Pyth. theorem)

∴ cm 6OC

3. In the figure, PQ and RS are tangents to the

circle at A and B respectively. Which of the

following is true?

在圖中，PQ 和 RS 分別是圓於 A 和 B 的

切線。下列哪一項是正確的？

A. x = a + b B. x = 180 – (a + b)

C. 2

bax

D.

290

bax

B

Question Bank

60

Join AB.

TAB = b ( in alt. segment)

ABT = a ( in alt. segment)

In △ABT,

)(180

180

180

180

ba

bax

axb

ABTATBTAB

( sum of △)

4. In the figure, the circle with centre C touches

△PQR at X, Y and Z. Find QPR.

在圖中，△PQR 與圓心為 C 的圓相切於

X、Y 和 Z。求 QPR。

A. 80

B. 90

C. 100

D. 110

A

100

134126360

360 XCYZCXZCY (s at a pt.)

PZC = PYC = 90 (tangent radius)

∵

80

3601009090

360

QPR

QPR

ZCYPYCPZCQPR

5. In the figure, AB is the tangent to the circle at

P. Find x.

在圖中，AB 是圓於 P 的切線。求 x。

A. 69

B. 71

C. 73

D. 75

D

57

180123

180

SPQ

SPQ

SRQSPQ (opp. s, cyclic quad.)

x

SPAPQS

( in alt. segment)

In △PQS,

75

1805748

180

x

x

SPQQSPPQS ( sum of △)

6. In the figure, BA and CD are tangents to the

circle at A and D respectively. Find ABC.

在圖中，BA 和 CD 分別是圓於 A 和 D

的切線。求 ABC。

A. 120

B. 130

C. 140

D. 150

C

Draw point E on the major arc

AD of the circle. Join

AE and ED.

70

BADAED ( in alt. segment)

70

AEDADC ( in alt. segment)

140

360807070

360

ABC

ABC

ABCDCBADCBAD

7. In the figure, CD is the tangent to the circle at

D, BAD = 34, DB = DC and CBA is a

straight line. Find BDA.

在圖中，CD 是圓於 D 的切線。

BAD = 34、DB = DC 及 CBA 是一線直

線。求 BDA。

A. 36

B. 37

C. 38

D. 39


61

D

34

BADBDC ( in alt. segment)

In △BCD,

CDBD

∴ DCBDBC (base s, isos. △)

180234

180

DBC

DCBDBCBDC ( sum of △)

73

2

34180DBC

In △ADB,

39

7334

BDA

BDA

DBCBDADAB (ext. of △)

8. In the figure, BC, CA and AB are tangents to

the circle at P, Q and R respectively. Which

of the following must be true?

A. BP = PC

B. RQ // BC

C. RB + QC = BC

D. A, R, P and Q

are concyclic.

在圖中，BC、CA 和 AB 分別與圓相切於 P、Q

和 R。下列哪一項是正確的？

A. BP = PC

B. RQ // BC

C. RB + QC = BC

D. A、R、P 和 Q

共圓。

C

BR = BP (tangent properties)

PC = CQ (tangent properties)

∵ BP + PC = BC

∴ BR + QC = BC

9. In the figure, the two circles touch at P as

shown and SP is their common tangent. SQ is

the tangent to the smaller circle at Q and SR

is the tangent to the larger circle at R. Find

PQR.

在圖中，兩個圓接觸於 P 點，而 SP 是它

們的公共切線。SQ 是較小的圓於 Q 的切

線，而 SR 則是較大的圓於 R 的切線。求

PQR。

A. 153

B. 154

C. 156

D. 158

A

SP = SQ (tangent properties)

SPQ = SQP (base s, isos. △)

In △SPQ,

71

2

38180

18038

SQP

SQPSPQ ( sum of △)

∵ SP = SQ (tangent properties)

SP = SR (tangent properties)

∴ SQ = SR

SQR = SRQ (base s, isos. △)

In △SQR,

82

2

16180

18016

SQR

SRQSQR ( sum of △)

∴

153

7182PQR

10. In the figure, PQ is the tangent to the circle at

A and ABC is a straight line. Which of the

following is true?

在圖中，PQ 是圓於 A 的切線且 ABC 是

一條直線。下列哪一項是正確的？

A. y = x B. y = 2x

C. x + y = 180 D. x + 2y = 180

A

Question Bank

62

Join BE.

x

EAQEBA

( in alt. segment)

∵ CDE = EBA (ext. , cyclic quad.)

11. In the figure, PB and PC are tangents to the

circle at A and C respectively and BOC is a

straight line. Which of the following is / are

true?

在圖中，PB 和 PC 分別是圓於 A 和 C

的切線，而 BOC 是一條直線。下列何者

為正確？

I. △BAO △BCP

II. BC = CP

III. O, A, P and C are concyclic.

O、A、P 和 C 共圓。

A. I only B. I and II only

C. II and III only D. I and III only

A. 只有 I

B. 只有 I 及 II

C. 只有 II 及 III

D. 只有 I 及 III

D

BAD = 90 tangent radius

BCP = 90 tangent radius

∴ BAO = BCP

ABO = PBC common angle

∴ △BAO ~ △BCP AAA

∴ I is true.

OAP = 90 tangent radius

OCP = 90 tangent radius

OAP + OCP

= 90 + 90

= 180

∴ O, A, P and C are concyclic. opp. s supp.

∴ III is true.

12. In the figure, ABP, ACR and BQC are

tangents to the circle at P, R and Q

respectively. Which of the following is / are

true?

在圖中，ABP、ACR 和 BQC 分別與圓相

切於 P、R 和 Q。下列何者為正確？

I. AB = AC

II. AB + BQ = AC + CQ

III. BP + CR = BC

A. I only B. II only

C. III only D. II and III only

A. 只有 I B. 只有 II

C. 只有 III D. 只有 II 及 III

D

∵ AP = AR (tangent properties)

∴ AB + BP = AC + CR

∵ BQ = BP (tangent properties)

and CQ = CR (tangent properties)

∴ AB + BQ = AC + CQ

∴ II is true.

BC

QCBQCRBP

∴ III is also true.

∴ y = x

13 In the figure, O is the centre of the circle.

AOBP is a straight line and PC is the tangent

to the circle at C. Express y in terms of x.

在圖中，O 是圓的圓心。AOBP 是一條直

線，而 PC 則是該圓於 C 的切線。試以 y

表示 x。

A. 2

45x

y

B. y = 45 + x

C. 2

90x

y

D. y = 90 – x


63

A

Join CB.

ACBACDBCP

ABC

180

90

90180 y

y 90

ACDABC y

In △BCD,

ABCBCPBPC

yyx )90(

902 xy

452

xy


A. Which of the following is true?

在圖中，PQ 是圓於 A 的切線。下列哪一

項是正確的？

A. x = a + b – c

B. x = a + c – b

C. x = 180 – a – b + c

D. x = 180 – a – b – c

D

ABC = x ( in alt. segment)

CAD = b (s in the same segment) In △ABC,

cbax

xbac

ABCCABACB

180

180)(

180 ( su

15. In the figure, BF AD, EB AC and

AD CE. ABC, AFD and CDE are straight

lines. Which of the following is / are true?

在圖中，BF AD，EB AC 及 AD CE。

ABC、AFD 和 CDE 都是直線。下列何者

為正確？

I. B, G, D and C are concyclic.

II. B, F, D and C are concyclic.

III. A, B, D and E are concyclic.

I. B、G、D 和 C 共圓。

II. B、F、D 和 C 共圓。

III. A、B、D 和 E 共圓。

A. I only

B. II only

C. I and III only

D. II and III only

A. 只有 I

B. 只有 II

C. 只有 I 及 III

D. 只有 II 及 III

C

180

9090

GDCCBG

∴ B, G, D and C are concyclic. opp. s supp.

∴ I is true.

EBC = ADC = 90 given

∴ ABE = ADE = 90 adj. s on st. line

∴ A, B, D and E are concyclic. converse of s in

the same segment

∴ III is true.

16. The figure shows three circles ABDH, BCGF

and CDHG. ABCD and EFGH are straight

lines. Which of the following is / are true?

圖中所示為三個圓 ABDH、 BCGF 和

CDHG。ABCD 和 EFGH 都是直線。下列

何者為正確？

I. ABFE is a cyclic quadrilateral.

II. ACGE is a cyclic quadrilateral.

III. BDHF is a cyclic quadrilateral.

I. ABFE 是一個圓內接四邊形。

Question Bank

64

II. ACGE 是一個圓內接四邊形。

III. BDHF 是一個圓內接四邊形。

A. I only

B. II only

C. III only

D. I, II and III

A. 只有 I

B. 只有 II

C. 只有 III

D. I、II 及 III

A

BCG = EHD ext. , cyclic quad.

BFE = BCG ext. , cyclic quad.

∴ BFE = EHD

EAD + EHD = 180 opp. , cyclic quad.

∵ EAD = EAB

∴ EAB + EHD = 180

i.e. EAB + BFE = 180

∴ ABFE is a cyclic quadrilateral. opp. s supp.

∴ I is true.

17. In the figure, TP is the tangent to the circle at

P and RP is a diameter of the circle. Find

QPT.

在圖中，TP 是圓於 P 的切線，而 RP 是

圓的一條直徑。求 QPT。

A. 36

B. 42

C. 48

D. 54

B

48

QSRQPR


∵ RP is a diameter of the circle.

∴ RPT = 90 (tangent properties)

42

9048

QPT

QPT

RPTQPTQPR

18. In the figure, CA and CE are tangents to the

circle at A and D respectively. AKD, BKE and

ABC are straight lines. Find KED.

在圖中，CA 和 CE 分別是圓於 A 和 D

的切線。AKD、BKE 和 ABC 都是直線。

求 KED。

A. 35

B. 36

C. 37

D. 38

C

In △AKB,

81

55136

ABKAKEKAB (ext. of △)

CA = CD (tangent properties)

81

DACADC (base s, isos. △)

99

81180

180 ADCKDE (adj. s on st. line)

In △EDK,

37

13699

KED

KED

AKEKEDKDE (ext. of △)

19. In the figure, TA is the tangent to the circle at

A and CBT is a straight line. If BA = BT and

BAT = 38, find BAC.

在圖中，TA 是圓於 A 的切線，而 CBT 是

一條直線。若 BA = BT 及 BAT = 38，求

BAC。

A. 48

B. 52

C. 58

D. 66

D

∵ AB = BT

∴

38

BATATB (base s, isos. △)

104

3838180

180 BATATBABT (ext. of △)

38

BATACB ( in alt. segment)

In △ABC,

66

10438

BAC

BAC

ABTACBBAC (ext. of △)


65

20. In the figure, TP and TQ are tangents to the

circle at P and Q respectively. If PTQ = 72,

find PRQ.

在圖中，TP 和 TQ 分別為圓於 P 和 Q

的切線。若 PTQ = 72，求 PRQ。

A. 48

B. 54

C. 63

D. 67

B

Join PQ.

TP = TQ (tangent properties)

TPQ = TQP (base s, isos. △)

In △TPQ,

54

2

72180

18072

TQP

TQPTPQ ( sum of △)

54

TQPPRQ ( in alt. segment)

21. In the figure, TB is the tangent to the circle at

A. O is the centre of the circle, and TCOD is a

straight line. If ATC = ADC, which of the

following is / are true?

在圖中，TB 是圓於 A 的切線。O 是該圓

的圓心，而 TCOD 是一條直線。若

ATC = ADC，下列何者為正確？

I. △ABD △CAD

II. △ABD △DBT

III. △TAC △CDA

A. I only B. II only

C. I and II only D. I and III only

A. 只有 I B. 只有 II

C. 只有 I 及 II D. 只有 I 及 III

C

ABD

CAD

90 in semi-circle

DAB = DCA in alt. segment

∴ △ABD ~ △CAD AAA

∴ I is true.

DBT

ABD

90

∵ △ABD ~ △CAD

∴ ADB = CDA

∵ CDA = DTB given

∴ ADB = DTB

∴ △ABD ~ △DTB AAA

∴ II is true.


A and TCB is a straight line. Which of the

following must be true?

在圖中，TA 是圓於 A 的切線，而 TCB 是

一條直線。下列何者必為正確？

I. △TAC △ABC

II. △TBA △ABC

III. △TAC △TBA

A. I only

B. II only

C. III only

D. I, II and III

A. 只有 I

B. 只有 II

C. 只有 III

D. I、II 及 III

C

TAC = TBA in alt. segment

ATC = BTA common angle

∴ △TAC ~ △TBA AAA

∴ III is true.

Question Bank

66

23. In the figure, CD is a diameter of the circle.

TS is the tangent to the circle at A. If

CDA = 48, find DAT.

在圖中，CD 是圓的一條直徑。TS 是圓於

A 的切線，若 CDA = 48，求 DAT。

A. 42

B. 44

C. 46

D. 48

.

A

Join AC.

DAC = 90 ( in semi-circle)

In ACD,

42

9048180

180 DACCDADCA ( sum of △)

42

DCADAT ( in alt. segment)

24. In the figure, O is the centre of the circle. TD

is the tangent to the circle at C. DEFG and

AGOBT are straight lines. If DG AT and

FAG = 26, find EDC.

在圖中，O 是圓的圓心。TD 是該圓於 C

的切線。DEFG 和 AGOBT 都是直線。若

DG AT 和 FAG = 26，求 EDC。

A. 52

B. 54

C. 56

D. 58

A

In △AFG,

64

2690180AFG ( sum of △)

64

AFGDFC (vert. opp. s)

Join CB.

ACB = 90 ( in semi-circle)

26

CABBCT ( in alt. segment)

In △CDF,

52

269064

EDC

EDC

ACTDFCEDC (ext. of △)

25. In the figure, TP and TQ are tangents to the

circle at A and B respectively. ATB = 72

and CA = CB. Find CBQ.

在圖中，TP 和 TQ 分別是圓於 A 和 B

的切線，ATB = 72 及 CA = CB.，求

CBQ。

A. 63

B. 66

C. 68

D. 72

A


TAB = TBA (base s, isos. △)

In ATB,

54

180272

180

TBA

TBA

TBATABATB ( sum of △)

54

TBAACB ( in alt. segment)

∵ CA = CB

∴ CAB = CBA (base s, isos. △)

In △ ABC,

63

180254

180

CAB

CAB

CBACABACB ( sum of △)

63

CABCBQ ( in alt. segment)


67


C. AC is a diameter of the circle, ABP and

ADQ are straight lines. If AQC = 57, find

ABD.

在圖中，PQ 是圓於 C 的切線。AC 是圓

的一條直徑，而 ABP 和 ADQ 都是直線。

若 AQC =

57 ，求

ABD。

A. 54

B. 57

C. 62

D. 64

B

Join CD.

ADC = 90 ( in semi-circle)

In CDQ,

33

9057

DCQ

DCQ (ext. of △)

OCQ = 90 (tangent radius)

57

3390

DCQOCQACD

57

ACDABD (s in the same segment)

27. In the figure, the three sides of △ABC touch

the circle at the points D, E and F. If

ABC = 68 and ACB = 54, find

DEFD : .

在圖中，△ABC 的三邊分別與圓相切於

D、E 和 F。若 ABC = 68 及

ACB = 54，求

DEFD : 。

A. 5 : 6

B. 6 : 7

C. 7 : 8

D. 8 : 9

D

Let O be the centre of the circle and join FO, EO and

DO.

OEC = ODC = 90 (tangent radius)

126

549090360EOD

BFO = BDO = 90 (tangent radius)

112

689090360FOD

9

8

126

112

EOD

FOD

DE

FD

(arcs prop. to s at centre)

∴ 9:8:

DEFD

28. In the figure, PQ and RS are tangents to the

circle at A and C respectively. AC and BD

intersect at K. If PAB = 47 and

DCS = 39, find AKD.

在圖中，PQ 和 RS 分別是圓於 A 和 C

的切線。AC 和 BD 相交於 K。若

PAB = 47 及 DCS = 39，求 AKD。

A. 82

B. 86

C. 94

D. 98

C

47

PABBCA ( in alt. segment)

39

DCSDBC ( in alt. segment)

Question Bank

68

In BCK,

94

3947180

180 DBCBCABKC ( sum of △)

94

BKCAKD (vert. opp. s)


A and BC is a diameter of the circle. If

QPC = 22 and ABC = 35, find PCB.

在圖中，PQ 是圓於 A 的切線，而 BC 是

圓的一條直徑。若 QPC = 22 及

ABC = 35，求 PCB。

A. 42

B. 44

C. 47

D. 49

A

35

ABCCAQ ( in alt. segment)

In △APC,

13

3522

PCA

PCA

CAQPCAAPC (ext. of △)

CAB = 90 ( in semi-circle)

In △ABC,

42

18090)13(35

180

PCB

PCB

CABBCAABC ( sum of △

30. In the figure, PB is the tangent to the circle at

B. COP is a straight line and AB // CP. If

ABC = 24, find BPO.

在圖中，PB 是圓於 B 的切線。COP 是一

條直線，而 AB // CP。若 ABC = 24，求

BPO。

A. 36

B. 39

C. 42

D. 45

C

24

ABCBCO (alt. s, CP // AB)

48

224BOP ( at centre twice at ☉ce)

OBP = 90 (tangent radius)

In △BOP,

42

1804890

180

BPO

BPO

BOPOBPBPO ( sum of △)

31. In the figure, AP and AQ are tangents to the

larger circle at P and Q respectively. The

smaller circle is inscribed in △ABC. BC is a

common tangent to both circles. If the

perimeter of △ABC = 27 cm, find AP.

在圖中，AP 和 AQ 分別是較大的圓於 P

和 Q 的切線，而較小的圓則內接於

△ABC。BC 是兩個圓的公共切線。若

△ABC 的周界 = 27 cm，求 AP。

A. 9 cm

B. 12 cm

C. 13.5 cm

D. 15 cm

C

With the notation in the figure,

BP = BX (tangent properties)

CQ = CX (tangent properties)

AP = AQ (tangent properties)

Perimeter of △ABC

AP

AQAP

CQACBPAB

ACCQBPAB

ACXCBXAB

ACBCAB

2

)()(

)(

)(

∴

cm 5.13

2

cm 27

AP


69

32. In the figure, the circle is inscribed in the

quadrilateral ABCD. If AD = 7 cm,

DC = 10 cm and BC = 15 cm, find AB.

圖中的圓內接於四邊形 ABCD。若

AD = 7 cm，DC = 10 cm 及 BC = 15 cm，

求 AB。

A. 11 cm

B. 12 cm

C. 13 cm

D. 14 cm

B

AP

APADPD

AQAP

cm 7


AP

DPDS

cm 7


AP

AP

DSDCSC

cm 3

)cm 7(cm 10

AP

SCRC

cm 3


AP

AP

RCBCBR

cm 12

)cm 3(cm 15

AP

BRQB

cm 12


cm 12

)cm 12(

APAP

QBAQAB

33. In the figure, the three sides of △ABC touch

the circle at the points D, E and F. If

AB = 20 cm, BC = 14 cm and CA = 18 cm,

find BD.

在圖中，△ABC 的三邊分別與圓相接於

D、E 和 F。若 AB = 20 cm，BC = 14 cm 及

CA = 18 cm，求 BD。

A. 7 cm

B. 8 cm

C. 9 cm

D. 10 cm

B

AF = AE (tangent properties)

FB = BD (tangent properties)

cm 8

cm )1826(

cm 26

cm 26

cm 52)(2

cm 5222

cm 52

cm 522

cm 52)()(

cm 52(

cm )181420(

CABD

CABD

CABD

CABD

CA2BDCA

CABDECAE

CAECBDBDAE

CADC)(BDFB)AF

CABCAB

34. In the figure, PA and PB are tangents to the

circle at A and B respectively. O is the centre

of the circle and COA is a straight line. If

BAC = 21, find BCA.

在圖中，PA 和 PB 分別是圓於 A 和 B

的切線。O 是該圓的圓心，而 COA 是一

條直線。若 BAC = 21，求 BCA。

A. 42

B. 45

C. 48

D. 49

C

OAP = 90 (tangent radius)

69

2190

BACOAPBAP

PB = PA (tangent properties)

69

PABPBA (base s, isos. △)

In △ABC,

48

6921

BCA

BCA

PBABCACAB (ext. of △)


A and 3:4:5::

CABCAB . Find ATC.

在圖中，TA 是圓於 A 的切線，而

3:4:5::

CABCAB 。求 ATC。

A. 25

B. 30

Question Bank

70

C. 35

D. 40

B

∵ 3:4:5:: CABCAB

∴ ACB : BAC : ABC = 5 : 4 : 3

(arcs prop. to s at ☉ce)

Let ABC = 3k, BAC = 4k and ACB = 5k.

In △ABC,

15

180543

180

k

kkk

ACBBACABC ( sum of △)

∴ ABC = 45, BAC = 60 and ACB = 75

45

CBACAT ( in alt. segment)

In △ABT,

30

180)4560(45

180

ATC

ATC

BATTBAATC ( sum of △)

basic properties of circles (ii) · 7 basic properties of circles (ii) 3 5. in the figure, oab is a...

Documents