2 Basic Properties of Circles - WordPress.com · 05.08.2014 · 2 Basic Properties of Circles 27...

2 Basic Properties of Circles

27


Review Exercise 2 (p. 2.7)

1. (a)

°=

°=°+°+°+

75

3606513090

x

x (∠s at a pt.)

(b) )// s, (corr.60 RSPQCBP ∠°=∠

°=

∠∠=

60

s) opp. vert.( CBPy

2. (a)

°=∠

°=°+∠

55

180125

PQR

PQR (adj. ∠s on st. line)

°=

°=°+

∠=∠+

85

14055

x

x

PRBPQRx (ext. ∠ of △)

(b)

°=

°=°+°+

50

1809040

a

a (∠ sum of △)

°=

°+°=

°+=

84

3450

34ab (ext. ∠ of △)

3. Yes, EF is parallel to GH.

∠RSQ = 110° (corr. ∠s, AB // CD)

∵ °=°+°=∠+∠ 18011070RSQPQS

∴ EF // GH (int. ∠s supp.)

4. ∠BCD = 65° alt. ∠s, AB // CD

°=°+∠+∠ 18050BDCBCD ∠ sum of △

°=

°−°−°=∠

65

6550180BDC

∵ ∠BCD = ∠BDC = 65°

∴ BD = BC sides opp. equal ∠s

∴ △BCD is an isosceles

triangle.

5. ∵ △PQR is an equilateral triangle.

∴ ∠PRQ = 60°

∵ PR is the median of QS in △PQS.

∴ RS = QR = PR

∴ ∠RPS = ∠RSP (base ∠s, isos. △)

°=∠

°=∠

∠=∠+∠

30

602

RPS

RPS

PRQRSPRPS (ext. ∠ of △)

6. (a) In △ACE and △DCB,

2

5

2

32

=

+=

DC

AC

2

5

3

5.7

3

5.52

=

=

+=

CB

CE

∴ CB

CE

DC

AC=

∠ACE = ∠DCB common angle

∴ △ACE ~ △DCB ratio of 2 sides, inc. ∠

(b) ∵ △ACE ~ △DCB proved in (a)

∴

cm 25.6

cm 2

cm 5

cm 5.2

=

=

=

AE

AE

DC

AC

DB

AE

corr. sides, ~ △s

7. (a) In △ABC,

∵ ∠C = 90°

∴

cm 10

cm 5.75.12

theorem)(Pyth.

22

222

=

−=

=+

AC

ABBCAC

(b) In △CAD,

2

22222

cm 676

cm )2410(

=

+=+ ADAC

2

222

cm 676

cm 26

=

=CD

∵ AC2 + AD2 = CD2

∴ ∠CAD is a right angle. (converse of Pyth.

theorem)

(c) Area of quadrilateral ABCD

= area of △ABC + area of △CAD

2

2

2

cm 5.157

cm )1205.37(

cm 10242

1105.7

2

1

=

+=

××+××=

8. (a) In △ABC and △ACD,

°=∠=∠ 90CDABCA given

DACCAB ∠=∠ common angle

CABBCAABC ∠−∠−°=∠ 180 ∠ sum of △

DACCDA ∠−∠−°= 180 proved

ACD∠= ∠ sum of △

∴ △ABC ~ △ACD AAA

(b) (i) In △ABC,

∵ ∠ACB = 90°

∴

cm 17

cm 158 22

222

=

+=

+=

AB

BCACAB (Pyth. theorem)

NSS Mathematics in Action 5A Full Solutions

28

(ii) ∵ △ABC ~ △ACD

∴

cm 17

64

cm 8

cm 8

cm 17

=

=

=

AD

AD

AD

AC

AC

AB(corr. sides, ~ △s)

fig.) sig. 3 to(cor. cm 2.13

cm 17

6417

=

−=

−= ADABBD

9. (a) In △ADE and △CDE,

sidecommon

given

given 90

DEDE

CDAD

DECDEA

=

=

°=∠=∠

∴ △ADE ≅ △CDE RHS

(b) Yes, △ABD and △AED are congruent.

°=∠

°=°+°+∠

°=°+∠+∠

30

1809060

18090

ACB

ACB

BACACB (∠ sum of △)

∵ △ADE ≅ △CDE (proved in (a))

∴

°=

∠=∠

30

ECDEAD (corr. ∠s, ≅ △s)

∴

°=

°−°=∠

30

3060BAD

In △ABD and △AED,

sidecommon

proved 30

given 90

ADAD

EADBAD

AEDABD

=

°=∠=∠

°=∠=∠

∴ △ABD ≅ △AED AAS

Activity

Activity 2.1 (p. 2.22)

2. (b) 2=∠

∠

APB

AOB

(c) No matter where points B and P are,

∠AOB = 2∠APB.

3. (b) 2Reflex

=∠

∠

AQB

AOB

(c) No matter where points B and Q are,

reflex ∠AOB = 2∠AQB.


1. (a) ∠AOB (i.e. c) is the angle at the centre subtended by

arc AB.

(b) ∠APB and ∠AQB are the angles at the circumference

in the same segment. They are both subtended by arc

AB.

2. (a) ∠APB = 2

c

( ∠ at centre twice ∠ at ⊙ce )

(b) ∠AQB =

2

c

( ∠ at centre twice ∠ at ⊙ce )

3. ∠APB = ∠AQB


1. (a) Yes (b) Yes (c) Yes

2. Yes


1. (b) ∠A + ∠C = 180° , ∠B + ∠D = 180°

2. ∠P + ∠R = 180° , ∠Q + ∠S = 180°

3. The sum of the opposite angles of a cyclic quadrilateral is

180°.

Activity 2.5 (p. 2. 53)

1. (a) Yes (b) Yes

2.

3. Yes

Classwork

Classwork (p. 2.11)

1. Element Term

AB • • minor arc

region BEC • • major arc

)

AFB • • chord

region OBFC • • major sector

region AFCEB • • major segment

)AB • • minor segment

region OBEC • • minor sector

2. (a) The two circles with the same centre are

concentric circles .

(b) The circle is the circumcircle of △ABC.

(c) The circle is the inscribed circle of △PQR.


29

Classwork (p. 2.24)

(a)

°=

°×=

∠=

60

302

2 APBx (∠ at centre twice ∠ at ⊙ce)

(b)

°=

°×=

∠=

50

1002

1

2

1AOBx

(∠ at centre twice ∠ at ⊙ce

)

(c) Reflex

°=

°−°=∠

220

140360AOB (∠s at a pt.)

°=

°×=

∠=

110

2202

1

reflex 2

1AOBx

(∠ at centre twice ∠ at ⊙ce)

Classwork (p. 2.25)

1. (a) °= 90x (∠ in semi-circle)

(b) ∠APB = 90° (∠ in semi-circle)

°=

°=+°+°

°=+°+∠

50

1804090

18040

x

x

xAPB (∠ sum of △)

2. OA = OP (radii)

°=∠

∠=∠

55OPA

OPAOAP

(base ∠s, isos. △)

°=°+°=∠ 903555APB

∴ The line segment joining A and B is a diameter of the

circle. (converse of ∠ in semi-circle)

Classwork (p. 2.36)

(a) ∵ °=∠=∠ 43AOBDOC (given)

∴

4=

=

x

ABDC))

(equal ∠s, equal arcs)

(b) ∵ ))ABCD = (given)

∴ ABCD = (equal arcs, equal chords) ∴ 5=x

(c) ∵ AB = DC (given)

∴ °= 65x (equal chords, equal ∠s)

Classwork (p. 2.39)

(a)

32

)80(5

2

=

°=°

=∠

°

x

x

AB

BC

AOB

x)

) (arcs prop. to ∠s at centre)

3

80

485

cm

=

°

°=

∠

∠=

y

AOB

DOC

AB

y) (arcs prop. to ∠s at centre)

(b)

10

30

506

cm

=

°

°=

∠

∠=

x

AOB

DOC

AB

x) (arcs prop. to ∠s at centre)

°=

°−°−°=

∠−∠−°=∠

100

5030180

180 DOCAOBBOC (adj. ∠s on st. line)

20

30

1006

cm

=

°

°=

∠

∠=

y

AOB

BOC

AB

y) (arcs prop. to ∠s at centre)

(c)

5.7

40

3010

cm

=

°

°=

∠

∠=

x

BEC

CED

BC

x) (arcs prop. to ∠s at ⊙ce)

24

)40(10

6

=

°=°

=∠

°

y

y

BC

AB

BEC

y)

) (arcs prop. to ∠s at ⊙ce)

(d)

°=

°=∠

=∠

∠

36

)48(4

3ACB

CD

AB

DBC

ACB)


96

3648180

180

=

°−°−°=

∠−∠−°=°

x

ACBDBCx (∠ sum of △)

Classwork (p. 2.47)

(a)

°=

°=°+

°=∠+∠

120

18060

180

x

x

BCDDAB (opp. ∠s, cyclic quad.)

°=

°=°+

°=∠+∠

100

18080

180

y

y

CDAABC (opp. ∠s, cyclic quad.)


30

(b)

°=

°=+°

°=∠+∠

87

18093

180

x

x

BCDDAB (opp. ∠s, cyclic quad.)

°=

∠=∠

113y

CDEABC (ext. ∠, cyclic quad.)

(c)

°=

°+°=

78

3246y (ext. ∠ of △)

°=

=

78

yx (ext. ∠, cyclic quad.)

Classwork (p. 2.55)

1. (a) ∵

°≠

°=

°+°=∠+∠

180

170

70100BCDBAD

∴ A, B, C and D are not concyclic.

(b)

°=∠

°=°+∠

°=∠+∠

80

180100

180

ABC

ABC

FBCABC (adj. ∠s on st. line)

∵ ADEABC ∠≠∠

∴ A, B, C and D are not concyclic.

(c)

°=∠

°=°+°+∠

20

18012040

ABD

ABD (∠ sum of △)

°=

°+°+°+°=∠+∠

180

)7040()5020(ADCABC

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

2. (a) ∵ ∠PSQ =∠PRQ = 50° given

∴ PQRS is a cyclic

quadrilateral. converse of ∠s in the

same segment

°=

°=°+°+

°=∠+∠

60

180)7050(

180

x

x

RSPPQR (opp. ∠s, cyclic quad.)

(b)

°=

°+°=∠

90

6030QPS

∴ ∠QRK = ∠QPS = 90°

∴ PQRS is a cyclic

quadrilateral. ext. ∠ = int. opp. ∠

°=∠

°=°+°+∠

°=∠+∠+∠

50

1809040

180

PRQ

PRQ

QRKPRSPRQ

(adj. ∠s on st. line)

°=

∠=∠

50x

PRQPSQ

(∠s in the same segment)

Quick Practice

Quick Practice 2.1 (p. 2.13)

Join OQ.

OQ = 13 cm (radius)

In △OQN, ∵ ON 2 + NQ2 = OQ2 (Pyth. theorem) ∴

cm 12

cm 513 22

22

=

−=

−= ONOQNQ

∵ ON ⊥ PQ (given) ∴ PN = NQ (line from centre ⊥ chord bisects chord)

cm 24

cm )12(2

2

=

=

=

+=

NQ

NQPNPQ

Quick Practice 2.2 (p. 2.14) ∵ PN = QN (given) ∴ ON ⊥ PQ (line joining centre to mid-pt. of

chord ⊥ chord)

∠ONP = 90°

In △OPN, ∵ ON = PN ∴ ∠OPN = ∠PON (base ∠s, isos. △)

°=∠

°=°+∠

°=∠+∠+∠

45

180902

180

OPN

OPN

ONPPONOPN (∠ sum of △)

Quick Practice 2.3 (p. 2.14) ∵ PN = NR (given) ∴ ON ⊥ PR (line joining centre to mid-pt. of

chord ⊥ chord) ∴ △OPN is a right-angled triangle.

Let r cm be the radius of the circle. ∴ OP = r cm and ON = (r − 1) cm

In △OPN,

13

262

2512

5)1(

22

222

222

=

=

++−=

+−=

+=

r

r

rrr

rr

PNONOP (Pyth. theorem)

∴ The radius of the circle is 13 cm.

cm 12

cm )113(

=

−=ON

Quick Practice 2.4 (p. 2.17) ∵ M and N are the mid-points of PQ and RQ respectively. ∴ OM ⊥ PQ and ON ⊥ RQ (line joining centre to mid-pt.

of chord ⊥ chord)

cm 6

cm )33(

=

+=

+= MQPMPQ

∵ ON = OM (given) ∴ RQ = PQ (chords equidistant from centre are equal)

= 6 cm


31

∴

cm 3

cm 62

1

2

1

=

×=

= RQRN


(a) ∵ AB = CD, OM ⊥ AB given

and ON ⊥ CD

∴ OM = ON equal chords, equidistant

from centre

∠MPN = ∠APC = 90° vert. opp. ∠s

∴ ∠MON = 90° ∠sum of polygon

∵ All four interior angles are equal to 90° and two

adjacent sides are equal.

∴ ONPM is a square.

(b) ∵ ONPM is a square. (proved in (a))

∴ PM = OM = 4 cm

∵ OM ⊥ AB (given) ∴ AM = MB (lines from centre ⊥ chord bisects

chord)

∴

cm 3

cm )411(cm 4

=

−=+

−=+

AP

AP

PMPBPMAP


°=

∠°−°=∠

242

pt.) aat s( 118360reflex AOC ∴ °=

°=

∠=

121

2422

reflex 2

x

x

AOCx (∠ at centre twice ∠ at ⊙ce)

∴ °=

°=

∠=

59

1182

2

y

y

AOCy (∠ at centre twice ∠ at ⊙ce)


∠CAD = 90° (∠ in semi-circle) ∵ AB = AD (given) ∴ ∠ABD = ∠ADB = y (base ∠s, isos. △)

In △ABD,

°=

°=

°=°+°++

°=∠+∠+∠

25

502

180)4090(

180

y

y

yy

BADADBABD

(∠ sum of △)

In △ACD,

°=

°=°++°

°=∠+∠+∠

65

1802590

180

x

x

ADCACDCAD

(∠ sum of △)


∠DOC = 2∠DBC ∠ at centre twice ∠ at ⊙ce

∠ODB = ∠DBC alt. ∠s, OD // BC

In △ODP,

°=∠

°=∠

°=∠+∠

°=∠+∠

30

903

902

90

DBC

DBC

DBCDBC

DOPODP

ext. ∠ of △

°=

°+°=

∠+°=∠

90

3060

60 DBCABC

∴ AC is a diameter of the circle. converse of ∠ in

semi-circle


°=

∠=∠

32

BDCBAC


°=∠ 90BCD (∠ in semi-circle)

In △ABC,

°=

°−°−°=∠

28

32120180BCA (∠ sum of △) ∴

°=

°−°=∠

62

2890ACD


x

DABDEB

=

∠=∠ (∠s in the same segment)

∠AEB = 90° (∠ in semi-circle)

In △ACE,

°=

°=

°=°+

°=+°+°++°

°=∠+∠+∠

8

162

1801642

180)90()50(24

180

x

x

x

xx

AECCAEECA (∠ sum of △)


Join OB, OC, OE, OF and OG.

Let ∠AOB = x. ∵ AB = BC = CD = DE = EF = FG = GH (given) ∴ ∠AOB = ∠BOC = ∠COD = ∠DOE =

∠EOF = ∠FOG = ∠GOH = x (equal chords, equal ∠s)

°=

°=

°=∠+∠+∠

40

1203

120

x

x

CODBOCAOB

°=∠

°=∠+°

°=∠+

∠°=∠+∠+∠+

∠+∠+∠+∠+∠

80

360)40(7

3607

pt.) aat s( 360

AOH

AOH

AOHx

AOHGOHFOG

EOFDOECODBOCAOB


32

Quick Practice 2.12 (p. 2.37) ∵ OP = OQ = OR, OP ⊥ AB,

OQ ⊥ BC and OR ⊥ AC given ∴ AB = BC = AC chords equidistant from

centre are equal ∴ )))ACBCAB == equal chords, equal arcs

Quick Practice 2.13 (p.2.40)

Let ∠COD = x. ∵ BC = CD (given) ∴ ∠BOC =∠COD = x (equal chords, equal ∠s)

)

)

BC

AD

BOC

AOD=

∠

∠ (arcs prop. to ∠s at centre)

xAOD7

10=∠

°=

°=

°=+++°

°=∠+∠+∠+∠

5.66

2287

24

3607

10132

360

x

x

xxx

DOACODBOCAOB (∠s at a pt.)


xACB

CD

AB

DBC

ACB

2

3=∠

=∠

∠)


In △BCK,

°=

°=

°=+

∠=∠+∠

30

752

5

752

3

x

x

xx

AKBCBKBCK (ext. ∠ of △)

Quick Practice 2.15 (p.2.48) Join DC. ∵ AD = AB (given) ∴

x

ABDADB

=

∠=∠ (base ∠s, isos. △)

∠BDC = 90° (∠ in semi-circle)

°=

°=

°=°+

°=°++°+

°=∠+∠

34

682

1801122

180)90()22(

180

x

x

x

xx

ADCABC

(opp. ∠s, cyclic quad.)


∠FAD = x (ext. ∠, cyclic quad.)

In △FAD,

xADF

xADF

FADDFAADF

−°=∠

°=+°+∠

°=∠+∠+∠

137

18043

180 (∠ sum of △)

x

ADFEDC

−°=

∠=∠

137

(vert. opp. ∠s)

In △DCE,

°=

°=

=−°+°

∠=∠+∠

86

1722

)137(35

x

x

xx

BCDEDCCED (ext. ∠ of △)


(a) ∵

°≠

°=

°+°=∠+∠

180

210

100110FEBBAF

∴ A, B, E and F are not concyclic.

(b) ∵

°=

°+°=∠

110

2090CDF

∴ BEFCDF

BEF

∠≠∠

°=∠ 100

∴ F, E, C and D are not concyclic.

(c) ∵ ∠ADB = 20° and ∠ACB = 20°

∴ ∠ADB = ∠ACB

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

same segment)

Quick Practice 2.18 (p.2.57) ∵ ∠FAB + ∠BCD = 180° int. ∠s, AF // CD

and ∠BEF = ∠BCD ext. ∠, cyclic quad. ∴ ∠FAB + ∠BEF = 180° ∴ A, B, E and F are concyclic. opp. ∠s supp.


(a) ∵ △ABC is an equilateral triangle.

∴ ∠ABC = 60° prop. of equil. △

∵ △CDF is an equilateral triangle.

∴ ∠DFC = 60° prop. of equil. △

∵ ∠ABC = ∠DFC

∵ B, C, E and F are concyclic. converse of ∠s in

the same segment

(b) ∵ B, C, E and F are concyclic. (proved in (a))

∴

°=

∠=∠

33

BFCBEC (∠s in the same segment)

∵ ∠BAC = 60° (prop. of equil. △)

∴

°=

°−°=

∠−∠=∠

27

3360

BECBACFCE (ext. ∠ of △)


33

Further Practice

Further Practice (p. 2.15)

1. (a) ∵ CN = ND (given)

∴ ON ⊥ CD (line joining centre to mid-pt. of

chord ⊥ chord)

∴ ∠ONC = 90°

In △OND,

°=

°−°=

∠−∠=∠

55

3590

ODNONCNOD (ext. ∠ of △)

(b) In △OME,

cm 8

cm 61022

22

222

=

−=

−=

=+

OMOEEM

OEEMOM (Pyth. theorem)

∵ OM ⊥ EF (given)

∴

cm 8=

= EMMF (line from centre ⊥ chord bisects chord)

2. ∵ CD is the perpendicular bisector of the chord AB.

∴ CD is a diameter of the circle.

cm 20

cm )515(

=

+=

+= MDCMCD

Radius of the circle

cm 10

cm 202

1

2

1

=

×=

= CD

Let O be the centre of the circle.

cm 5

cm )510(

cm 10

=

−=

−=

=

MDODOM

OA

In △OAM,

cm) 310(or cm 752

2

cm 75

cm 510

theorem)(Pyth.

22

22

222

=

=

+=

=

−=

−=

+=

AM

MBAMAB

OMOAAM

AMOMOA


1. (a) ∵ ON = OM = 4 cm, OM ⊥ AB, ON ⊥ CD (given)

∴ CD = AB (chords equidistant from

= 7 cm centre are equal)

cm 5.3

cm 72

1

=

×=

= NDCN (given)

(b)

cm 4=

= APPB (given)

∴ OP ⊥ AB (line joining centre to mid-pt.

of chord ⊥ chord)

Also, OQ = OP

and OQ ⊥ BC (given)

∴ BC = AB (chords equidistant from

= 8 cm centre are equal)

cm 4

2

1

2

1

=

=

=

=

AB

BCQC

BQQC (line from centre ⊥ chord bisects chord)

(c) ∵ OM ⊥ AB, ON ⊥ CD (given)

∴ MB = AM (line from centre ⊥ chord

bisects chord)

∴

cm10

cm 52

AB

=

×=

and

cm10

cm 52

CD

=

×=

∵ AB = CD, OM ⊥ AB and ON ⊥ CD

∴ ONOM = (equal chords, equidistant

cm 5.2= from centre)

2.

Construct OM and ON such that OM ⊥ PQ and ON ⊥ RS.

Note that MON is a straight line, since PQ // RS.

cm 8

cm 162

1

2

1

=

×=

=

=

PQ

MQPM (line from centre ⊥ chord bisects chord)

∵ PQ = RS, OM ⊥ PQ and ON ⊥ RS

∴

cm 6

cm 122

1

=

×=

= ONOM

centre) from

t equidistan chords, (equal

In △PMO,

cm 10

cm 68 22

22

222

=

+=

+=

+=

OMPMOP

OMPMOP (Pyth. theorem)

∴ Radius of the circle cm 10=


34


1. (a) Consider △ABD.

°=∠

°=°+°+°+∠

°=∠+∠+∠

50

18070)2040(

180

ABD

ABD

ADBBADABD (∠ sum of △)

°=

∠=

50

ABDx (∠s in the same segment)

(b)

°=

∠=∠

48

BDCBAC


∠ABC = 90° (∠ in semi-circle)

In △ABC,

°=

°=+°+°

°=∠+∠+∠

42

1809048

180

x

x

BCAABCBAC (∠ sum of △)

2.

°=

∠=∠

40

BCDBOD (opp. ∠s of // gram)

°=

°×=

∠=∠

20

402

1

2

1BODBAD (∠ at centre twice ∠ at ⊙ce

)

∴ °=

∠=∠

20p

BADADO

(alt. ∠s, OD // AC)

3. ∠ADC = 90° (∠ in semi-circle)

In △APD,

°=∠

°=∠+°+°+°

°=∠+∠+∠

24

18090)3630(

180

APD

APD

APDPDAPAD (∠ sum of △)

∴ °=∠ 24BPC


1.

Join OB.

∵ AB = BC = DE (given)

∴ ∠AOB = ∠BOC = ∠DOE (equal chords, equal ∠s)

= 55°

∵ ))

CDAE = (given)

∴ ∠AOE = ∠COD (equal arcs, equal ∠s)

°=∠

°=∠

°=∠+°+∠+°+°

°=∠+∠

+∠+∠+∠

5.97

1952

360555555

360

COD

COD

CODCOD

EOADOE

CODBOCAOB (∠s at a pt.)

2.

Join PR.

°=∠ 90PRS (∠ in semi-circle)

1

3

21

=

+=

=∠

∠))

RS

PR

SPR

PSR (arcs prop. to ∠s at ⊙ce)

∴ SPRPSR ∠=∠

°=∠

°−°=∠

°=∠+∠+∠

45

901802

180

PSR

PSR

PRSSPRPSR

(∠ sum of △)


1. (a)

°=

∠=∠

81

ABCDEC (ext. ∠, cyclic quad.)

°=

°−°−°=

∠−∠−°=∠

34

6581180

180

x

ECDDECCDE (∠ sum of △)

(b)

Join OD.

∵ OD = OA (radii)

∴

°=

∠=∠

18

OADODA (base ∠s, isos. △)

∵ OD = OC (radii)

∴

°=

∠=∠

50

OCDODC (base ∠s, isos. △)

°=

°+°=∠

68

5018ADC

°=

°=°+

°=∠+∠

112

18068

180

x

x

ADCABC (opp. ∠s, cyclic quad.)

2. ∵ BC = CD (given)

∴ ))

CDBC = (equal chords, equal arcs)

)

)

CD

BC

CAD

BAC=

∠

∠ (arcs prop. to ∠s at ⊙ce)

128

=°

∠BAC

°=∠

°=∠

90

28

ACB

BAC

(∠ in semi-circle)


35

°=

°−°−°=

∠−∠−°=∠

62

9028180

180 ACBBACABC

(∠ sum of △)

°=∠

°=°+∠

°=∠+∠

118

18062

180

ADC

ADC

ABCADC


Exercise

Exercise 2A (p. 2.18)

Level 1

1. ∵ ON ⊥ AB (given)

∴

cm 8

cm 162

1

chord) bisects chord centre from (line

=

×=

⊥= ANBN

Join OB.

Consider △NOB.

cm 10

cm 68 22

22

=

+=

+= ONBNOB

(Pyth. Theorem)


2.

Join OB.

cm 13

cm )85(

=

+=

+=

=

NCON

OCOB

(radii)

∵ AN = NB (given)

∴ ON ⊥ AB (line joining centre to mid-pt. of

chord ⊥ chord)

Consider △ONB.

cm 12

cm 513 22

22

=

−=

−= ONOBNB

(Pyth. theorem)

∵ AN = NB

∴

cm 24

cm 122

=

×=AB

3. ∵ AM = MB (line from centre ⊥ chord bisects chord)

∴

cm 12

cm 62

=

×=AB ∵ ON = OM, OM ⊥ AB and ON ⊥ CD (given)

∴

cm 12

equal) are centre fromt equidistan (chords

=

= ABCD

4. ∵ CN = ND (given)

∴ ON ⊥ CD (line joining centre to mid-pt. of chord

⊥ chord)

∴ ∠ONK = 90°

∵ AM = MB (given)

∴ OM ⊥ AB (line joining centre to mid-pt. of chord

⊥ chord)

∴ ∠OMK = 90°

°=

°−°=

∠−°=∠

137

43180

180 DKBMKN


Consider quadrilateral OMKN.

°×−=

∠+∠+∠+∠

180)24(

MKNONKOMKMON

(∠ sum of polygon)

°=

°−°−°−°=∠

43

1379090360MON

5. ∵ AM = MB (given)

∴ OM ⊥ AB (line joining centre to mid-pt. of chord

⊥ chord)

cm 3

cm 62

1

=

×=MB

Consider △OMB.

cm 4

cm 3522

22

=

−=

−= MBOBOM

(Pyth. theorem)

cm 3

cm )47(

=

−=

−= OMMNON

Join OD.

Consider △OND.

cm 5=

= OBOD

(radii)

cm 4

cm 35 22

22

=

−=

−= ONODND

(Pyth. theorem)

∵ ON ⊥ CD (given)

∴ CN = ND (line from centre ⊥ chord bisects

chord)

∴

cm 8

cm 42

=

×=CD


36

6. ∴ CM = MD (given)

∵ OM ⊥ CD (line joining centre to mid-pt. of chord

⊥ chord)

∴ ∠OMC = 90°

°=

°+°=

∠+∠=∠

122

9032

OMCOCMAOC

(ext. ∠ of △)

Consider △OAC.

∵ OC = OA (given)

∴ ∠OCA = ∠OAC (base ∠s, isos. △)

°=∠

°=°+∠

°=∠+∠+∠

29

1801222

180

OCA

OCA

AOCOACOCA

7. ∵ OM ⊥ AP (given)

∴ AM = MP (line from centre ⊥ chord bisects

chord)

∴

cm 12

cm 242

1

2

1

=

×=

= APMP

Join OP.

cm 15

cm 129 22

22

=

+=

+= MPOMOP

(Pyth. theorem)

∵ ON ⊥ BP (given)

∴ PN = NB (line from centre ⊥ chord bisects

chord)

∴

cm 9

cm 182

1

2

1

=

×=

= BPPN

Consider △ONP.

cm 12

cm 915 22

22

=

−=

−= PNOPON

(Pyth. theorem)

8.

cm 10

cm )416(2

1

)(2

1

2

1

=

+=

+=

=

MBAM

ABOB

∴

cm 6

cm )410(

=

−=

−= MBOBOM

Join OD.

Consider △OMD.

cm 10=

= OBOD

(radii)

cm 8

cm 610 22

22

=

−=

−= OMODMD

(Pyth. theorem)

∵ OM ⊥ CD (given)

∴ CM = MD (line from centre ⊥ chord bisects

chord)

∴

cm 16

cm 82

2

=

×=

= MDCD

9. (a) ∵ PQ is the perpendicular bisector of the chord RS.

∴ PQ is a diameter of the circle.

∵ PQ = 10 cm (given)

∴ radius of the circle

cm 5

cm 102

1

2

1

=

×=

= PQ

Join OR.

OR = 5 cm (radius)

cm 4

cm 35 22

22

=

−=

−= TROROT

(Pyth. theorem)

OP = 5 cm (radius)

∴

cm 9

cm )45(

=

+=

+= OTOPPT

(b)

cm 1

cm )910(

=

−=

−= PTPQTQ


37

10. ∵ BM = MC = 6 cm (given)

∴ OM ⊥ BC (line joining centre to mid-pt. of chord

⊥ chord)

Consider △OMB.

cm 8

cm 610 22

22

=

−=

−= BMOBOM

(Pyth. theorem)

Consider △OMD.

cm 15

cm 817 22

22

=

−=

−= OMODMD

(Pyth. theorem)

∴

cm 9

cm )615(

=

−=

−= MCMDCD

11. Construct a circle with centre O lying on BH, such that the

circle cuts AB at two points P and Q, and cuts BC at two

points R and S as shown.

Draw OM and ON such that OM ⊥ AB and ON ⊥ BC.

∠ABH = ∠CBH given

∠OMB = ∠ONB = 90° by construction

OB = OB common side

∴ △OBM ≅ △OBN AAS

∴ OM = ON corr. sides, ≅ △s

∴ PQ = RS chords equidistant from centre

are equal

Level 2

12.

Construct OMN such that OM ⊥ CD and ON ⊥ AB.

OMN is a straight line.

OA = OC = 17 cm (radii) ∵ OM ⊥ CD (by construction) ∴

cm 15

cm 302

1

2

1


=

×=

=

⊥=

CD

MDCM

In △OCM,

cm 8

cm 1517 22

22

=

−=

−= CMOCOM

(Pyth. theorem)

∵ ON ⊥ AB (constructed) ∴

cm 8

cm 162

1

2

1


=

×=

=

⊥=

AB

NBAN

In △OAN,

cm 15

cm 817 22

22

=

−=

−= ANOAON

(Pyth. theorem)

∴ Distance between AB and CD

cm 7

cm )815(

=

−=

−= OMON

13.

Let M be a point on AB such that OM ⊥ AB.

∵ OM ⊥ AB (by construction)

∴

cm 12

cm 242

1


=

×=

⊥= MBAM

Consider △OMA.

cm 9

cm 121522

22

=

−=

−= AMOAOM

(Pyth. theorem)

Consider △OMC.

cm 40

cm )2812(

=

+=

+= BCMBMC

cm 41

cm 40922

22

=

+=

+= MCOMOC

(Pyth. theorem)

cm 26

cm )1541(

=

−=

−= ODOCCD

14. (a) Consider △ABP and △ACP.

AP = AP common side

BP = CP given

OP ⊥ BC line joining centre to

mid-pt.of chord ⊥ chord

∴ ∠APB = ∠APC

= 90°

∴ △ABP ≅ △ACP SAS

∵ AB = AC corr. sides, ≅ △s

∴ △ABC is an isosceles triangle.


38

(b) ∵ OM ⊥ AB (given)

∴

cm 12

cm 62

chord) bisects2

chord centre from (line

=

×=

=

⊥=

AMAB

MBAM

cm 12=

= ABAC

(proved in (a))

cm 12

cm 62

2

=

×=

= BPBC (given)

∵ AC = BC, ON ⊥ AC and OP ⊥ BC

∴

cm 32

centre) fromt equidistan chords, (equal

=

= ONOP

15. (a) Consider △OAB and △OAC.

OA = OA common side

OB = OC radii

AB = AC given

∴ △OAB ≅ △OAC SSS

∴ ∠OAB = ∠OAC corr. ∠s, ≅ △s

∴ AO bisects ∠BAC.

(b) Consider △ABN and △CAN.

AB = AC given

∠OAB = ∠OAC proved in (a)

AN = AN common side

∴ △ABN ≅△CAN SAS

∴ BN = CN corr. sides, ≅ △s

∴ ON ⊥ BC line joining centre to

mid-pt. of chord ⊥ chord

(c)

cm 3

cm )58(

=

−=

−= OAANON

Consider △ONC.

cm 4

cm 3522

22

=

−=

−= ONOCNC

(Pyth. theorem)

Consider △ANC.

)cm 54(or cm 80

cm 48 22

22

=

+=

+= NCANAC (Pyth. theorem)

16. (a)

cm )3( −=

−=

r

NYOYON

(b) ∵ ON ⊥ AB (given)

∴ NBAN = (line from centre ⊥ chord

cm 9

cm 182

1

=

×=

bisects chord)

Consider △OAN.

OA = r cm (radius)

∴ 15

906

9)3(

22

222

222

=

=+−

=+−

=+

r

rrr

rr

OAANON (Pyth. theorem)

17. ∵ OM ⊥ CD (given)

∴ MDCM = (line from centre ⊥ chord

cm 6

cm 122

1

=

×=

bisects chord)

Let r cm be the radius of the circle.

cm )18( r

OAAMOM

−=

−=

Join OC.

Consider △OCM.

OC = r cm (radius)

10

36360

6)18(

22

222

222

=

=+−

=+−

=+

r

rrr

rr

OCCMOM (Pyth. theorem)

∴

cm 2

cm 18)10(2

cm 18)(2

cm )]18([

=

−×=

−=

−−=

−=

r

rr

OMOBMB

18.

Let M be a point on AB such that OM ⊥ AB. ∵ OM ⊥ AB (by construction)

∴ AMMB = (line from centre ⊥ chord

cm 9

cm 182

1

=

×=

bisects chord)

Join OB.

OB = 13 cm (radius)

Consider △OMB.

cm 88

cm 913 22

22

=

−=

−= MBOBOM

(Pyth. theorem)

Let N be a point on CD such that ON ⊥ CD.

∵ ON ⊥ CD (by construction)

∴

cm 12

cm 242

1


=

×=

⊥= DNNC

∵

°=

∠=∠

90

OMKONK

∴ ONKM is a rectangle.


39

∴ NK = OM (property of rectangle)

∴

d.p.) 2 to(cor. cm 2.62

cm )8812(

=

−=

−=

−=

OMNC

NKNCKC

19. (a)

cm 6

cm )51(

=

+=

+= RQPRPQ

cm 3

cm 62

1

2

1

=

×=

= PQOP

cm 2

cm )13(

=

−=

−= PROPOR

In △ORM,

cm 3

cm 12 22

22

=

−=

−= OMORRM

(Pyth. theorem)

∵ OM ⊥ RS (given)

∴ RM = MS (line from centre ⊥ chord bisects

chord)

∴

cm 32

2

=

= RMRS

(b)

Join OD.

OD = OP = 3 cm (radii)

In △OND,

cm 8

cm 13 22

22

=

−=

−= ONODND

(Pyth. theorem)

∵ ON ⊥ CD (given) ∴ CN = ND (line from centre ⊥ chord bisects chord) ∴

cm) 24(or cm 82

2

=

= NDCD

20. (a) ∵ MCMN2

1= ∴ NCMN =

∵ MN = NC and BD ⊥ MC

∴ BD is the perpendicular bisector of chord MC.

∴ BD is a diameter of the circle.

(b) (i)

cm 8

cm 162

1

2

1

=

×=

= MCNC

In △BCN,

cm 6

cm 81022

22

=

−=

−= NCBCNB

(Pyth. theorem)

Join OC.

Let r cm be the radius of the circle.

cm r

OBOC

=

=

(radii)

cm )6( −=

−=

r

NBOBON

In △OCN,

d.p.) 2 to(cor. 33.8

12

100

10012

643612

8)6(

22

222

222

=

=

=

++−=

+−=

+=

r

r

r

rrr

rr

NC ON OC (Pyth. theorem)

∴ The radius of the circle is 8.33 cm.

(ii)

cm 3

50

cm 12

1002

2

=

×=

= OBBD

cm 13.33

cm 103

50 2

2

22

=

−

=

−= ADBDAB

(Pyth. theorem)

Exercise 2B (p. 2.30)

Level 1

1.

°=∠

°=°+∠

42

180138

ACB

ACB


°=

°×=

∠=

84

422

2 ACBx


2. ∠ACB = 90° (∠ in semi-circle)

∵ CA = CB (given)

∴ x = ∠CBA (base ∠s, isos. △)

°=

°=+°

°=+∠+∠

45

180290

180

x

x

xCBAACB (∠ sum of △)


40

3.

°=

∠=∠

55

ABDACD


°=

°=°+

°=∠+

70

12555

125

x

x

ACDx

(ext. ∠ of △)

4.

°=

∠=∠

25

DBEDAC


°=

°+°=

∠+∠=

67

4225

ADCDACx

(ext. ∠ of △)

5.

°=

°×=

∠=∠

106

532

2 ACBAOB


°=

°=°+

°=∠+

12

118106

118

x

x

AOBx

(ext. ∠ of △)

6.

°=

°−°=

∠−∠=∠

58

3290

CADCDBACD

(ext. ∠ of △)

∠ACB = 90° (∠ in semi-circle)

∴

°=

°−°=

∠−∠=

32

5890

ACDACBx

7. °=∠ 90ADC (∠ in semi-circle)

°=

°−°−°=

∠−∠−°=∠

25

6590180

180 DACADCDCA

(∠ sum of △)

°=

∠=

25

DCAx


8.

x

x

ACBAOB

4

)2(2

2

=

=

∠=∠ (∠ at centre twice ∠ at ⊙ce)

∵ OB = OA (radii)

∴ ∠OBA = x (base ∠s, isos. △)

°=

°=

°=++

°=∠+∠+

30

1806

1804

180

x

x

xxx

AOBOBAx

(∠ sum of △)

9.

Join OC.

°=

°×=

∠=∠

65

1302

1

2

1AOBACB


∵ OC = OA (radii) ∴

°=

∠=∠

20

OACOCA (base ∠s, isos. △) ∵ OB = OC (radii) ∴

°=

°−°=

∠−∠=

∠=

45

2065

OCAACB

OCBx


10. ∠DAC = 90° (∠ in semi-circle)

°=

°−°−°=

∠−∠−°=∠

35

5590180

180 ACDDACADC

(∠ sum of △)

∵ AB = AD (given) ∴

°=

∠=∠

35

ADBABD


°=

°−°=∠

°=∠+∠

20

3555

55

BAC

BACABD

(ext. ∠ of △)

11.

(a)

°=

°−°=∠

°=∠+∠

148

32180

180

AOC

AOCOAB

(int. ∠s, BA // CO)

∴ Reflex

°=

°−°=

∠−°=∠

212

148360

360 AOCAOC

(∠s at a pt.)

(b)

°=

°×=

∠=∠

106

2122

1

reflex 2

1AOCABC


12. ∠BOD = 36° (opp. ∠s of // gram)

°=

°×=

∠=∠

18

362

1

2

1BODBCD


°=

∠=∠

18

BCDODC

(alt. ∠s, DO // AC)

°=

°+°=

∠+∠=∠

54

3618

BODODKBKD

(ext. ∠ of △)

13. (a) Reflex

°=

°×=

∠=∠

280

1402

2 ADCAOC


°=

°−°=∠

°=∠+∠

80

280360

360reflex

AOC

AOCAOC (∠s at a pt.)


41

(b)

Join OB. ∵ OB = OA (radii) ∴

°=

∠=∠

28

OABOBA


°=

°×=

∠=∠

40

802

1

2

1AOCABC


°=

°−°=

∠−∠=∠

12

2840

OBAABCOBC

∵ OC = OB (radii) ∴ °=

∠=∠

12

OBCBCO (base ∠s, isos. △)

14. ∠AEF, ∠AFE, ∠BDC, ∠DBC, ∠EAC and ∠CAF

(any four of the above angles)

15. Suppose ∠OBC = 2∠OAC.

°=

°×=

∠=∠

45

902

1

2

1AOBACB


OAC

OACAOKAKB

∠+°=

∠+∠=∠

90

(ext. ∠ of △)

OAC

OBCBCKBKA

∠+°=

∠+∠=∠

245

(ext. ∠ of △)

∴

°=∠

∠+°=∠+°

45

24590

OAC

OACOAC

If °=∠ 45OAC , then B and C will coincide, which

contradicts the assumption that B and C are distinct

points.

∴ ∠OBC ≠ 2∠OAC

(or any other reasonable answers)

Level 2

16.

°=

°−°=∠

°=∠+∠

22

2850

50

DCB

CKBDCB

(ext. ∠ of △)

°=

∠=∠

22

DCBDAB



°=∠

°=°+∠+°+°

°=∠+∠+∠

18

180)22(5090

180

CAD

CAD

CABCBAACB (∠ sum of △)

17. ∵ DC = DA (given)

∴ ∠DCA = x (base ∠s, isos. △)

∵ BD = BC (given)

∴

x

DCABDC

=

∠=∠


∠ADB = 90° (∠ in semi-circle)

°=

°=+°++

°=∠+∠+∠

30

180)90(

180

x

xxx

ADCDCADAC

(∠ sum of △)

∴ °=∠ 30BDC

18.

°=

∠=∠

23

DBCDAC


°=∠

°=∠+°+°

°=∠+∠+∠

90

1806723

180

ADC

ADC

ADCACDDAC

(∠ sum of △)

∴ AC is a diameter of the circle. (converse of ∠ in

semi-circle)

19. BCDBAD ∠=∠ (∠s in the same segment)

ABCBCD ∠=∠ (alt. ∠s, CD // AB) ∴ ABCBAD ∠=∠


°=∠

°=°+∠+∠

°=°+∠+°+∠

°=∠+∠+∠

23

180134

18090)44(

180

ABC

ABCABC

BADABC

ACBBACABC

(∠ sum of △)

20.

°=

°×=

∠=∠

62

1242

1

2

1AOEABE


°=

°−°=∠

∠=∠+∠

26

3662BEC

ABEBECACE

(ext. ∠ of △)

°=

∠=∠

26

BECBAD


∴

°=

°+°=

∠+∠=∠

88

6226

ABEBADAKE

(ext. ∠ of △)

21.

Join BD.

°=∠ 90ABD (∠ in semi-circle)

x

ABEABDEBD

−°=

∠−∠=∠

90

x

EBDECD

−°=

∠=∠

90



42

22. (a) OABC is a parallelogram. given

OA = OC radii

∴ OABC is a rhombus.

(b) Reflex xAOC −°=∠ 360 ∠s at a pt.

AOCABC ∠=∠ reflex 2

1 ∠ at centre twice

2180

)360(2

1

x

x

−°=

−°=

∠ at ⊙ce

(c) ∠ABC = x (opp. ∠s of // gram)

°=

°=

=−°

120

1802

3

2180

x

x

xx

(proved in (b))

23. (a) ∵ DK ⊥ BK

∴ BK = EK line from centre ⊥ chord

bisects chord

In △BKD and △EKD,

BK = EK proved

∠BKD = ∠EKD given

DK = DK common side

∴ △BKD ≅ △EKD SAS

(b) Let ∠BED = x.

x

KEDKBD

=

∠=∠ (corr. ∠s, ≅ △s)

x

BEDACEABE

+°=

∠+∠=∠

42

(ext. ∠ of △)

°=

°=++°

°=∠+∠

°=∠

24

9042

90

90

x

xx

KBDABE

ABD

(∠ in semi-circle)

∴

°=

∠=∠

24

BEDBAD


24. (a) In △AKB and △DKC,

∠AKB = ∠DKC vert. opp. ∠s

∠BAK = ∠CDK ∠s in the same segment

∠ABK = ∠DCK ∠s in the same segment

∴ △AKB ~ △DKC AAA

CK

DK

BK

AK= corr. sides, ~ △s

∴ DKBKCKAK •=•

(b) DKBKCKAK •=• (proved in (a))

DK•=• cm) (3cm) (2cm) 6(

∴ cm 4=DK

25.

Join AP.

∠APB = 90° ∠ in semi-circle

°=

∠=∠

90

APQACQ

∠s in the same segment

i.e. QC ⊥ AB

26.

Join OA.

ABP

AOQABQ

∠=

∠=∠

2

ceat twicecentreat

segment same in the s ⊙∠∠

∠

∴ BP bisects ∠ABQ.

27. (a) ∵ EA = EB given ∴ ∠EAB = ∠EBA base ∠s, isos. △

∠BAD = ∠BED ∠s in the same segment

EAD

BADEAB

BEDEBABCE

∠=

∠−∠=

∠−∠=∠

ext. ∠ of △

∴ ∠BCE = ∠EAD

(b) ∠ABE = ∠ADE (∠s, in the same segment)

∴

AED

DAEAED

CDA

ADE

ABECBE

∠>

∠∠+∠=

∠∠=

∠−°=

∠∠−°=∠

) of (ext.

line) st.on s (adj.

180

line) st.on s (adj. 180 △

∴ It is not possible that △BCE ~ △EAD.

Exercise 2C (p. 2.42)

Level 1

1. Reflex

°=

°−°=

∠−°=∠

280

80360

360 AOBAOB

(∠s at a pt.)

4

280

8014

Reflex Major

cm

=

°

°=

∠

∠=

x

AOB

AOB

AB

x)

(arcs prop. to ∠s at centre)

2.

°=

°−°−°=

∠−∠−°=∠

55

7550180

180 ACBABCBAC

(∠ sum of △)

11

50

5510

cm

=

°

°=

∠

∠=

x

ABC

BAC

AC

x) (arcs prop. to ∠s at ⊙ce

)


43

3. °=∠ 90ABC (∠ in semi-circle)

°−°=

°−°−°=

∠−∠−°=∠

x

x

ABCACBBAC

90

90180

180

(∠ sum of △)

50

4509

45450

5

490

=

°=°

°=°−°

=°

°−°

=∠

∠

x

x

xx

x

x

AB

CB

ACB

BAC))

(arcs prop. to ∠s at ⊙ce)

4.

°=

°+°=

∠+∠=∠

°=

°=∠

=∠

∠

140

5684

56

)84(6

4

CODBOCBOD

COD

BC

CD

BOC

COD))


70

70

1402

1

2

1

=

°=

°×=°

∠=∠

x

x

BODBAD


5. BADADC ∠=∠ alt. ∠s, CD // AB

∴ ))

BDAC : equal ∠s, equal arcs

6.

(a)

°=

°=∠

=∠

∠

64

)48(3

4AOB

BC

AB

BOC

AOB))


(b)

°=

°×=

∠=∠

32

642

1

2

1AOBACB


7.

°=

°−°−°=

∠−∠−°=∠

54

6462180

180 DCBDBCBDC

(∠ sum of △)

°=

°−°=

∠−∠=∠

27

3562

DEBDBCEDB (ext. ∠ of △) 2:1

54:27

)at s toprop. (arcs :: ce

=

°°=

∠∠∠= ⊙BDCEDBBCAB))

8. ∠BAC = 90° (∠ in semi-circle)

°=

°+°=

∠+∠=∠

120

3090

CADBACBAD

∵ AB = AD (given)

∴ ∠ABD = ∠ADB (base ∠s, isos. △)

°=∠

°=°+∠

°=∠+∠+∠

30

1801202

180

ABD

ABD

BADADBABD (∠ sum of △)

°=

°−°−°=

∠−∠−°=∠

60

3090180

180 ABCBACACB (∠ sum of △)

∴

3:2

90:60

)at s toprop. (arcs :: ce

=

°°=

∠∠∠= ⊙BACACBBCAB))

9. ∵ ))BAAC = (given)

∴

cm 6=

= BAAC

(equal arcs, equal chords)

°=∠ 90BAC (∠ in semi-circle)

In △ABC,

cm 72

cm 66 22

22

=

+=

+= ACABBC

(Pyth. theorem)

Perimeter of △ABC

fig.) sig. 3 to(cor. cm 20.5

cm )7266(

=

++=

10. ∵ CPD

BPCAPBAPC

∠=

°+°=

∠+∠=∠

105 ∴ 1=

∠

∠=

CPD

APC

CD

AC))


))

CDAC =

∵

EPF

CPDBPCBPD

∠=

°=

°+°=

∠+∠=∠

25

1510

∴ 1=∠

∠=

EPF

BPD

EF

BD))


))

EFBD =

∵

FPG

CPDBPCAPBAPD

∠=

°=

°+°+°=

∠+∠+∠=∠

30

15105

∴ 1=∠

∠=

EPG

APD

FG

AD))


))

FGAD = (equal ∠s, equal arcs)

∴ ))

CDAC = , ))

EFBD = , ))

FGAD =

(any two of the above answers)


44

11. ∵ 1==∠

∠))

CD

AB

DAC

ADB (arcs prop. to ∠s at ⊙ce)

∴ ∠ADB = ∠DAC

∴ KA = KD (sides opp. equal ∠s)

Similarly, KB = KC

∴ △AKD and △BKC are isosceles triangles.

∵ )))

CDBCAB == (given)

∴ AB = BC = CD (equal arcs, equal chords)

∴ △ABC and △BCD are isosceles triangles.

(any three of the above answers)

Level 2

12. ∠BAD = 90° (∠ in semi-circle)

°=

°−°=

∠−∠=∠

40

5090

CADBADBAC

12

50

4015

=

°

°=

∠

∠=

x

CAD

BAC

CD

BC))


13.

Join BC.

°=

∠=∠

70

BDABCA


°=

°−°−°=∠

°=∠+∠+∠

70

7040180

180

ABC

BCABACABC (∠ sum of △)

5.17

40

7010

=

°

°=

∠

∠=

x

BAC

ABC

BC

ADC))


14.

Let AC and BD intersect at E.

°=∠ 90BAD (∠ in semi-circle)

∵ AB = AD (given)

∴ ∠ABD = ∠ADB (base ∠s, isos. △)

°=∠

°=°+∠

°=∠+∠+∠

45

180902

180

ADB

ADB

BADADBABD

(∠ sum of △)

°=∠

°=∠+°+°

°=∠+∠+∠

60

1807545

180

DAE

DAE

DAEAEDADE

(∠ sum of △)

4

45

60

3

=

°

°=

∠

∠=

x

x

ADB

DAC

AB

CD))


15. (a) ∵ OD = OB (radii)

∴

°=

∠=∠

30

OBDODB


°=

°+°=

∠+∠=∠

60

3030

ODBOBDBOC

(ext. ∠ of △)

(b)

°=

°×=∠

=∠

∠

40

603

2AOB

CB

BA

BOC

AOB))


°=

°×=

∠=∠

20

402

1

2

1AOBADB

(∠ at centre twice ∠ at ⊙ce

)

16.

°=

∠=∠

30

PQSPRS


°=

°−°=

∠−∠=∠

45

3075

PRSQRSPRQ 1==

∠

∠))

QP

QR

PRQ

QSR (arcs prop. to ∠s at ⊙ce)

°=

∠=∠

45

PRQQSR °=

°−°−°=

∠−∠−°=∠

60

4575180

180 QSRQRSRQS

(∠ sum of △)

17. (a)


45

Join OB, OC and CD.

∵ )))CDBCAB == (given)

∴ ∠AOB = ∠BOC = ∠COD (equal arcs, equal ∠s)

°=∠

°=∠

°=∠+∠+∠

40

1203

120

BOC

BOC

CODBOCAOB

°=

°×=

∠=∠

20

402

1

2

1BOCBDC


(b) Reflex

°=

°−°=

∠∠−°=∠

240

120360

pt.) aat s(360 AODAOD

°=

°×=

∠∠∠×=∠

120

2402

1

)at twicecentreat ( reflex2

1 ce⊙AODACD

°=∠

°=°+°+∠

°=∠+∠+∠

40

18012020

180

CED

CED

ECDCDECED (∠ sum of △)

18. (a) (i)

Join OT, OP and OQ.

∵ RS = ST = TP = PQ = QR (given)

∠ROS = ∠SOT = ∠TOP =∠POQ = ∠QOR

(equal chords, equal ∠s)

∴

°=

°×=∠

72

3605

1ROS

(∠s at a pt.)

(ii)

°=

°×=

∠=∠

36

722

1

2

1ROSRPS


(b) Angle subtended at centre by any one side of the polygon

n

°=

360

Angle subtended at one of the vertices by any one side of

the polygon

n

n

°=

°×=

180

360

2

1

19. (a) OA = OC radii

AB = CB given

OB = OB common side

∴ △ABO ≅ △CBO SSS

(b) ∵ ∠AOB = ∠COB corr. ∠s, ≅ △s

∴

COB

AOBAOD

∠−°=

∠−°=∠

180

180

adj. ∠s on st. line

COD∠= adj. ∠s on st. line

∴ ))

DCAD = equal ∠s, equal arcs

20. ∵ 1==∠

∠))

CD

BC

DAC

CAB

arcs prop. to ∠s at ⊙ce

∴ ∠CAB = ∠DAC

∵ OC = OA radii

∴

DAC

CABACO

∠=

∠=∠

base ∠s, isos. △

∴ OC // AD alt. ∠s equal

21. (a) ∵ OE ⊥ BD given

∴ BE = ED line from centre ⊥ chord

bisects chord

AE = AE common side

∠AEB = ∠AED = 90° given

∴ △ABE ≅ △ADE SAS

(b) ∠BAC = ∠DAC corr. ∠s, ≅ △s

1=∠

∠=

DAC

BAC

CD

BC))

arcs prop. to ∠s at ⊙ce

∴ ))

CDBC =

22. (a) With the notations in the figure,

°=

°+°=

∠+∠=∠

50

3020

DBFBDFDFE

(ext. ∠ of △)

°=

°+°=

∠+∠=∠

90

5040

ACGCAGAGE

(ext. ∠ of △)

∴

°=

°−°−°=

∠−∠−°=

40

5090180

180 GFEFGEx

(∠ sum of △)

(b) )))))EADECDBCAB ::::

5:3:4:4:2

50:30:40:40:20

)at s toprop. (arcs

::::

ce

=

°°°°°=

∠

∠∠∠∠∠= ⊙ ACEDBECADBECADB

(c)

cm 9

cm 2

18circle theof nceCircumfere

53442

2

circle theof nceCircumfere

π

π

=

×=

++++=

)AB


46

∴ Radius of the circle

cm 4.5

cm 2

9

=

=π

π

23. If △ABC is a right-angled triangle, then either one of ∠A,

∠B or ∠C must be equal to 90°.

∠A : ∠B : ∠C = a : b : c (arcs prop. to ∠s at ⊙ce)

and ∠A + ∠B + ∠C = 180° (∠ sum of △)

∴ °×++

=∠ 180cba

aA ,

°×++

=∠ 180cba

bB

and °×++

=∠ 180cba

cC

Since one of ∠A, ∠B or ∠C is equal to 90°,

one of the ratios cba

b

cba

a

++++, or

cba

c

++is equal

to 2

1.

∴ The student’s claim is correct.


Exercise 2D (p.2.49)

Level 1

1. °=∠ 95BCD (ext. ∠, cyclic quad.)

°=

°−°=

°=+∠

85

95180

180

x

xBCD (adj. ∠s on st. line)

2.

°=

°−°=∠

°=∠+∠

104

76180

180

BCD

BADBCD (opp. ∠s, cyclic quad.)

∵ CD = CB (given)

∴ ∠BDC = x (base ∠s, isos. △)

∴

°=

°=°+

°=∠++∠

38

1801042

180

x

x

BDCxBCD (∠ sum of △)

3. ∠ACB = 90° (∠ in semi-circle)

°=

°−°−°=∠

∠°=∠+∠+∠

50

4090180

) of sum (180

ABC

BACACBABC △

°=

°−°=

°=∠+

130

50180

180

x

ABCx (opp. ∠s, cyclic quad.)

4.

°=

∠=

46

ABDx (∠s in the same segment)

°=

°+°−°=

°=+°+

∠°=∠+

80

)4654(180

180)54(

quad.) cyclic s, (opp.180

y

xy

BCDy

5.

°=

°−°=∠

°=∠+∠

70

110180

180

EBC

CDEEBC (opp. ∠s, cyclic quad.)

°=

°−°=∠

°=∠+∠

60

120180

180

ECB

BAEECB (opp. ∠s, cyclic quad.)

°=

°−°−°=∠

∠°=∠+∠+∠

50

6070180

) of sum (180

BEC

ECBEBCBEC △

6. (a)

°=

°+°=

∠+∠=∠

96

6036

ACDCADADE (ext. ∠ of △)

(b)

°=

∠=∠

60

BCDDFE (ext. ∠, cyclic quad.)

°=

°−°−°=∠

∠°=∠+∠+∠

24

9660180

) of sum (180

DEF

DEFFDEDFE △

7. ∠BAP = ∠PQD ext. ∠, cyclic quad.

∠DCP + ∠PQD = 180° opp. ∠s, cyclic quad.

∴ ∠DCP + ∠BAP = 180°

∴ AB // CD int. ∠s supp.

8.

°=

°−°=∠

°=∠+∠

50

130180

180

FCD

DEFFCD


°=

∠=

50

FCDx (ext. ∠, cyclic quad.)

∠ABD = y (ext. ∠, cyclic quad.)

∵ ABAD = (given)

∴

y

ABDADB

=


°=

°−°=

°=+

∠°=∠+∠+∠

65

501802

1802

) of sum (180

y

y

yx

ADBABDBAD △

9. (a)

°=

°×=

∠=∠

20

402

1

2

1AOBBPA (∠ at centre twice ∠ at ⊙ce

)

(b)

°=

°−°=∠

°=∠+∠

130

50180

180

BAP

BCPBAP (opp. ∠s, cyclic quad.)

°=

°−°−°=∠

∠°=∠+∠+∠

30

20130180

) of sum (180

ABP

APBBAPABP △


47

10.

°=

°−°−°=∠

∠°=∠+∠+∠

35

30115180

) of sum (180

DAC

ACDADCDAC △

∠ABC + ∠ADC = 180° (opp. ∠s, cyclic quad.)

°=

°−°=∠

65

115180ABC

∵ CDBC = (given) ∴ ))

CDBC = (equal chords, equal arcs) 1==∠

∠))

CD

BC

DAC

BAC (arcs prop. to ∠s at ⊙ce)

°=

∠=∠

35

DACBAC

°=

°−°−°=∠

∠°=∠+∠+∠

80

3565180

) of sum (180

ACB

BACABCACB △

11.

Join AD.

∠ABC + ∠CDA = 180° opp. ∠s, cyclic quad.

∠ADE = 90° ∠ in semi-circle

∠ABC + ∠CDE

°=

°+°=

∠+∠+∠=

∠+∠+∠=

270

90180

)(

)(

ADECDAABC

ADECDAABC

12. (a) Consider △ABC and

△ADC.

AB = AD given

BC = DC given

AC = AC common side

∴ △ABC ≅ △ADC SSS

(b) ∵ △ABC ≅ △ADC (proved in (a))

∴ ∠ABC = ∠ADC (corr. ∠s, ≅ △s)

°=∠

°=∠

°=∠+∠

90

1802

180

ABC

ABC

ADCABC


∴ AC is a diameter of the circle.

(converse of ∠ in semi-circle)

(c) If △ABC and △ADC are right-angled isosceles

triangles, then

°=

°+°=

∠+∠=∠

90

4545

DACBACBAD

and

°=

°+°=

∠+∠=∠

90

4545

DCABCABCD

∴ In this case, BD is a diameter of the circle.


13. ∵ △ABD is an equilateral triangle.

∴ ∠BAD = 60°

∵ △BCD is a right-angled triangle with ∠C = 90°.

∴ ∠BCD = 90°

∵

°≠

°=

°+°=∠+∠

180

150

9060BCDBAD

∴ The polygon cannot be a cyclic quadrilateral.


Level 2

14. Reflex

°=

°×=

∠=∠

220

1102

2 ABCCOA (∠ at centre twice ∠ at ⊙ce)

°=

°−°=

∠−°=∠

140

220360

reflex 360 COACOA (∠s at a pt.)

°=

°−°=∠

°=∠+∠

40

140180

180

APB

COAAPB (opp. ∠s, cyclic quad.)

15.

°=

∠=∠

40

DAECEB (corr. ∠s, EC // AD)

∵ EC = EB (given)

∴ EBCECB ∠=∠ (base ∠s, isos. △)

°=∠

°=∠+°

°=∠+∠+∠

70

180240

180

EBC

EBC

EBCECBCEB (∠ sum of △)

°=

°−°=∠

°=∠+∠

110

70180

180

ADC

EBCADC (opp. ∠s, cyclic quad.)

16. (a) ∠KAD = ∠KCB ext. ∠, cyclic quad.

∠KDA = ∠KBC ext. ∠, cyclic quad.

∠AKD = ∠CKB common angle

∴ △KAD ~ △KCB AAA

(b)

cm 1

cm 3cm 4

cm 4)(2

cm 3

cm 3

cm 2

=

+=

+=

+

+=

+

=

DC

DC

DC

ABKA

KD

DCKD

KA

KB

KD

KC

KA (corr. sides, ~ △s)

17.

°=

∠=∠

65

DCEDAB (ext. ∠, cyclic quad.)

∠ABD = 90° (∠ in semi-circle)

°=

°−°−°=∠

∠°=∠+∠+∠

25

6590180

) of sum (180

ADB

DABABDADB △

°=

∠=∠

25

ADBCBD (alt. ∠s, BC // AD)

°=

°−°=∠

∠=∠+∠

40

2565BDC

DCECBDBDC (ext. ∠ of △)


48

18.

ADC

ADCACD

CADADCACD

∠−°=

∠−°−°=∠

∠°=∠+∠+∠

140

40180

) of sum (180 △

ADCABC

ADCABC

∠−°=∠

°=∠+∠

180

180 (opp. ∠s, cyclic quad.)

°=∠+∠ 180ACDAED (opp. ∠s, cyclic quad.)

ADC

ADC

ACDAED

∠+°=

∠−°−°=

∠−°=∠

40

)140(180

180

∴

°=

∠+°+∠−°=∠+∠

220

)40()180( ADCADCAEDABC

19.

Join BD.

Let ∠CBE = x.

∵ CE = CB (given)

∴

x

CBECEB

=


°−=∠

∠=∠+∠

27xBEA

CBEBAEBEA

(ext. ∠of △)

°=∠ 90DBE (∠ in semi-circle)

°=

°=

°=+°−++°

°=∠+∠

39

1173

180)27()90(

180

x

x

xxx

DECDBC


∴ °=∠ 39CBE

20. Let ∠CO′D = x.

∠ABC = ∠CO′D = x ext. ∠, cyclic quad.

∵ OC = OB radii

∴

x

OBCOCB

=

∠=∠


∠OAD = ∠OCB = x ext. ∠, cyclic quad.

x

OBACDO

=

′∠=′∠

ext. ∠, cyclic quad.

∵ O′D = O′C radii

∴

x

CDODCO

=

′∠=′∠


°=

°=

°=′∠+′∠+′∠

60

1803

180

x

x

DOCCDODCO

∠ sum of △

∴ ∠O′DC = ∠O′CD = ∠CO′D = 60°

∴ △CDO′ is an equilateral triangle.

21. (a)

yx

yxx

PCBABCAPC

+=

++=

∠+∠=∠

2

)(

ext. ∠ of △

yx

xyy

RBCACBARB

2

)(

+=

++=

∠+∠=∠ ext. ∠ of △

(b) °=∠+∠ 180ARBAPC opp. ∠s, cyclic quad.

°=+

°=+

°=+++

60

18033

180)2()2(

yx

yx

yxyx

°=∠

°=°+∠

°=++∠

°=++∠

°=∠+∠+∠

60

180120

180)(2

18022

180

BAC

BAC

yxBAC

yxBAC

ACBABCBAC ∠ sum of △

Exercise 2E (p.2.58)

Level 1

1. (a) ∵ ∠BAD + ∠BCD

°=

°+°+°+°=

180

)6535()4238(

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

(b)

°=

∠=∠

38x

CABCDB (∠s in the same segment)

2. (a) ∵

°=

°−°−°=

∠−∠−°=∠

40

10535180

180 DCBBDCDBC ∠ sum of △

∴ ∠DBC = ∠DAC

∴ A, B, C and D are concyclic. converse of

∠s in the

same segment

(b)

°=

∠=∠

35x

BDCBAC (∠s in the same segment)

3. (a) ∵ ∠BAC = ∠CDB = 90°

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

in the same

segment

(b)

°=

°−°−°=

∠−∠−°=∠

35

9055180

180 CDBEDAADB (adj. ∠s on st. line)

∠ACB = ∠ADB (∠s in the same segment)

°= 35x

4. (a) Refer to the figure.

Consider △ABG.

°=

°−°−°=

∠−∠−°=∠

50

4585180

180 GABBGAABG ∠ sum of △

°=

°+°=

∠+∠=∠

85

3550

GBCABGABC

∴ ∠ABC = ∠ADF

∴ A, B, C and D are concyclic. ext.∠ = int.

opp. ∠


49

(b)

°=

∠=∠

35

CBDCAD (∠s in the same segment)

CADBAC

DABDCE

∠+∠=

∠=∠ (ext. ∠, cyclic quad.)

°=

°+°=

80

3545x

5. Consider △PBC.

°=∠

°=∠+°

∠=∠+∠

46

6216

PCB

PCB

APBPBCPCB ext. ∠ of △

°=

°−°=

∠−∠=∠

36

4682

PCBBCDACD

∴ ∠ABD = ∠ACD

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

the same segment

6. ∠ABC = ∠ADC opp. ∠s of // gram

∠PQB = ∠PDC ext. ∠, cyclic quad.

∠DPQ = ∠PQB alt. ∠s, AD // BC

∴ ∠DPQ = ∠ABC

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

7. ∵ AB = EB given

∴ ∠BAE = ∠BEA base ∠s, isos. △

∠BAE = ∠BCD opp. ∠s of // gram

∴ ∠BEA = ∠BCD

∴ B, C, D and E are concyclic. ext. ∠ = int. opp. ∠

8. ∠BPT = 90° ∠ in semi-circle

∠SQC = 90° ∠ in semi-circle

°=

°−°=

∠−°=∠

90

90180

180 SQCAQS adj. ∠s on st. line

∴ ∠AQR = ∠BPR = 90°

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

9. ∠BAD = 180° – (x + y) (∠ sum of △)

°=++−°

°=∠+∠

180)(180

180

zyx

BCDBAD (opp. ∠s, cyclic quad.)

zyx =+ (or any other equivalent

relationship)

10. ∵ AB = AC given

∴ ∠ABC = ∠ACB base ∠s, isos. △

∵ AM = MB and AN = NC given

∴ MN // BC mid-pt. theorem

∠AMN = ∠ABC corr. ∠s, MN // BC

∴ ∠AMN = ∠ACB

∴ B, C, N and M are concyclic. ext. ∠ = int. opp. ∠

11. D is any point on )AC , where AB is a diameter of the

circle and the mid-point of AB is the centre.

Level 2

12. (a) ∵ △ABC is an equilateral triangle.

∴ AB = BC = CA

Consider △ABD and △ACD.

AB = AC proved

BD = CD given

AD = AD common side

∴ △ABD ≅ △ACD SSS

Consider △ABE and △CBE.

AB = CB proved

AE = CE given

BE = BE common side

∴ △ABE ≅ △CBE SSS

(b) ∵ △ABD ≅ △ACD proved in (a)

∴ ∠ADB = ∠ADC corr. ∠s, ≅ △s

°=

∠=∠

°=∠+∠

90

180

ADCADB

ADCADB

adj. ∠s on st. line

Similarly, ∠BEA = ∠BEC = 90°

∴

°=

°+°=∠+∠

180

9090FECFDC

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

∵ ∠ADB = ∠BEA = 90°

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

the same segment

13. (a) ∠APB = 90° ∠ in semi-circle

°=∠

°=°+∠

°=∠+∠

90

18090

180

APM

APM

APBAPM adj. ∠s on st. line

°=∠ 90AOM given

∴ ∠APM = ∠AOM

∴ O, P, M and A are concyclic. converse of ∠s in

the same segment

(b) ∵ OPOA = radii ∴ ∠OAP = ∠OPA base ∠s, isos. △ ∠OAP = ∠OMP ∠s in the same segment

∴ ∠OPA = ∠OMB

14. ∠AQS = ∠BRS ext. ∠, cyclic quad.

∠BRS = ∠CPS ext. ∠, cyclic quad.

∴ ∠AQS = ∠CPS

∴ AQSP is a cyclic quadrilateral. ext. ∠ = int. opp. ∠


50

15. (a)

Join PB and let ∠ARP = θ.

∠APQ = ∠ARP = θ given

∠PBA = ∠ARP = θ ∠s in the same

segment

∠APB = 90° ∠ in semi-circle

θ−°=

∠−∠=∠

90

APQAPBBPQ

°=

−°−−°=

∠−∠−°=∠

90

)90(180

180

θθ

BPQPBQPQB ∠ sum of △

(b)

Join RB.

∠TQB = 90° proved in (a)

∠TRB = 90° ∠ in semi-circle

∠TQB + ∠TRB = 90° + 90° = 180°

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

16. Consider △ACB and △DBC.

AC = DB given

∠ACB = ∠DBC given

BC = CB common side ∴ △ACB ≅ △DBC SAS

∴ ∠BAC = ∠CDB corr. ∠s, ≅ △s

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

the same segment

17. (a) ∠ACB = 90° ∠ in semi-circle

°=

∠−°=∠

90

180 ACBECP adj. ∠s on st. line

°=∠ 90ADB ∠ in semi-circle

∴ ∠ECP = ∠ADB

∴ P, D, E and C are concyclic. ext. ∠ = int. opp. ∠

(b) ∠COD = 2∠CAD ∠ at centre twice

∠ at ⊙ce

APB

APBACPCAD

∠−°=

∠−∠−°=∠

90

180 ∠ sum of △

∴

APB

APBCOD

∠−°=

∠−°=∠

2180

)90(2

18. (a) ∵ ∠BAD + ∠BCD = 90° + 90°

= 180°

∴ BCDA is a cyclic quadrilateral. opp. ∠s supp.

∵ ∠BAD = 90° given

∴ BD is a diameter of the circle. converse of ∠s

in the same

segment

Since N is the mid-point of BD,

N is the centre of the circle.

∵ AP = PC given

∴ NP ⊥ AC line joining centre to mid-pt.

of chord ⊥ chord

(b)

Join MP and NP.

Let ∠ABD = x.

x

ABDACD

=

∠=∠ ∠s in the same segment

x

ACDBCDACB

−°=

∠−∠=∠

90

∵ AM = MB and AP = PC given

∴ MP // BC mid-pt. theorem

∴

x

ACBAPM

−°=

∠=∠

90

corr. ∠s, MP // BC

∴ NP ⊥ AC proved in (a)

∴ ∠APN = 90°

∴

ABD

x

x

APMAPN

NPM

∠=

=

−°−°=

∠−∠=

∠

)90(90

19. (a)

Join AE, BF, CG and DH.

EACCGE

CGEEAC

∠−°=∠

∠°=∠+∠

180

quad. cyclic s, opp.180

If A, E, F and B are concyclic,

then

EACBFE

BFEEAB

∠−°=∠

∠°=∠+∠

180

quad. cyclic s, opp.180

∴ ∠CGE = ∠BFE

∠BFE = ∠BDH ext. ∠, cyclic quad.

∴ ∠CDH = ∠CGE

∴ C, G, H and D are

concyclic. ext. ∠ = int. opp. ∠

(b)

Join PU, QT, RW and SV.

∠RSV = ∠PUT ext. ∠, cyclic quad.

∠VWR = ∠PQT ext. ∠, cyclic quad.

If R, S, W and V are

concyclic, then

∠RSV = ∠VWR ∠s in the same segment

∴ ∠PUT = ∠PQT

∴ P, Q, U and T are concyclic. converse of ∠s

in the same segment


51

(opp. ∠s,

cyclic quad.)

Revision Exercise 2 (p. 2.66)

Level 1

1.

Join OF.

Draw ON such that ON ⊥ FE.

cm 10

cm 202

1

2

1

=

×=

=

=

BC

OBOF (radii)

cm 6=

= ABON (property of rectangle)

Consider △ONF.

cm 8

cm 610 22

22

=

−=

−= ONOFFN (Pyth. theorem)

∵ ON ⊥ FE (by construction)

∴ FN = NE (line from centre ⊥

chord bisects chord)

∴

cm 16

cm 82

=

×=FE

2. ∠AEC = 90° (∠ in semi-circle)

°=

∠=∠

30

EACEBC (∠s in the same segment)

°=

°+°=

∠+∠=∠

65

3035

EBCEDCAEB (ext. ∠ of △)

°=

°−°=

∠−∠=∠

25

6590

AEBAECBEC

3. ∠POR + ∠OPQ = 180° (int. ∠s, OR // PQ)

°=

°−°=∠

138

42180POR

°=

°−°=

∠−°=∠

222

138360

360Reflex PORPOR (∠s at a pt.)

°=

°×=

∠=∠

111

2222

1

reflex 2

1PORPQR (∠ at centre twice ∠ at ⊙ce

)

°=

°−°=∠

°=∠+∠

69

111180

180

ORQ

PQRORQ (int. ∠s, OR // PQ)

4. With the notations in the figure,

°=

°×=

∠=∠

27

542

1

2

1AOBBCA (∠ at centre twice ∠ at ⊙ce

)

°=

°+°=

∠+∠=∠

69

2742

BCAOBCONC (ext. ∠ of △)

°=

°−°=∠

∠=∠+∠

15

5469OAC

ONCAOBOAC (ext. ∠ of △)

5. 1==∠

∠))

BC

AB

CAB

ACB (arcs prop. to ∠s at ⊙ce)

°=

∠=∠

35

CABACB

°=∠ 90DCA (∠ in semi-circle)

°=

°+°=

∠+∠=∠

125

3590

ACBDCADCB

°=∠

°=°+°+∠

°=∠+∠

20

180125)35(

180

DAC

DAC

DCBDAB (opp. ∠s, cyclic quad.)

6. EABEDC ∠=∠ (∠s in the same segment)

EAB

EDCADCADE

EAB

EABDABDAC

∠−°=

∠−∠=∠

∠−°=

∠−∠=∠

80

60

EAB

EABEAB

ADEDACAEB

∠−°=°

∠−°+∠−°=°

∠∠+∠=∠

2140100

)80()60(100

) of (ext. △

∴ °=∠ 20EAB

7. 1

2==

∠

∠))

CD

BC

CBD

CDB (arcs prop. to ∠s at ⊙ce)

∴ CDBCBD ∠=∠2

1

∴ °=∠

°=∠

°=∠

°=∠+∠+°

°=∠+°+∠+°

°=∠+∠

40

40

602

3

1802

1120

180)58()62(

180

KDC

CDB

CDB

CDBCDB

CDBCBD

ADCABC


52

8.

°=

°−°=∠

°=∠+∠

65

115180

180

ABC

ADCABC (opp. ∠s, cyclic quad.)

2

3=

=∠

∠))

BC

AB

BAC

ACB (arcs prop. to ∠s at ⊙ce)

∴ BACACB ∠=∠2

3

°=∠

°=∠

°=∠+°+∠

∠°=∠+∠+∠

46

1152

5

180652

3

) of sum (180

BAC

BAC

BACBAC

BACABCACB △

9. (a) ∵ OC = OB (radii) ∴ ∠OCB = ∠OBC (base∠s, isos. △)

°=∠

°=°+∠

°=∠+∠+∠

50

180802

180

OCB

OCB

BOCOBCOCB (∠ sum of △)

(b)

°=

°×=

∠=∠

40

802

1

2

1BOCBAC (∠ at centre twice ∠ at ⊙ce

)

quad.) cyclic s, (opp.180 ∠°=∠+∠ BCDDAB

°=∠

°=∠+°+°+°

°=∠+∠+∠+∠

54

180504036

180)()(

OCD

OCD

OCDOCBBACDAC

10.

Join BC.

°=

°×=

∠=∠

45

902

1

2

1DOCDBC (∠ at centre twice ∠ at ⊙ce

)

°=∠ 90ACB (∠ in semi-circle)

In △BCE,

°=∠

°=∠+°+°

∠°=∠+∠+∠

45

1804590

) of sum (180

CEB

CEB

CEBEBCBCE △

11.

Join DC.

∵ OC = OD (radii)

∴ ∠OCD = ∠ODC (base ∠s, isos. △)

°=∠+∠ 180BCDADC (int. ∠s, AD // BC)

°=∠

°=∠+°

°=∠+°+∠+°

58

180264

180)36()28(

OCD

OCD

OCDODC

∴

°=

°+°=∠

94

5836BCD

°=∠

°=°+∠

°=∠+∠

86

18094

180

BAD

BAD

BCDBAD (opp. ∠s, cyclic quad.)

12.

Draw OM such that OM ⊥ BC.

∵ OM ⊥ BC (by construction)

∴ BM = MC (line from centre ⊥ chord bisects chord)

∴

cm 3

cm 62

1

=

×=MC

Consider △OMC.

cm 4

cm 35 22

22

=

−=

−= MCOCOM (Pyth. theorem)

Consider △OAM.

cm 33

cm 47 22

22

=

−=

−= OMOAAM (Pyth. theorem)

∴

d.p.) 2 to(cor.cm 74.2

cm )333(

=

−=

−=

−=

MCAM

BMAMAB

13. (a) ∵ CF = FD (given)

∴ OF ⊥ CD (line joining centre to

mid-pt. of chord ⊥ chord)

cm 2

cm 42

1

2

1

=

×=

= CDFD

In △OFD,

cm 1

cm 2)5( 22

22

=

−=

−= FDODOF (Pyth. theorem)


53

(line from centre ⊥


(line from centre ⊥


(b)

Join AB.

∵ OE = OF = 1 cm (proved in (a))

and OE ⊥ AC and OF ⊥ CD (proved in (a))

∴

cm 4=

= CDAC

equal) are centre from

t equidistan (chords

∵ OE ⊥ AC (given)

∴

cm 2

cm 42

1

2

1

=

×=

=

=

AC

ECAE

cm 5=

= ODOB (radii)

cm )15( −=

−= OEOBBE

d.p.) 2 to(cor.cm 24.1

cm 2)15(2

1

2

1 of Area

2

2

=

×−×=

××= AEBEABE△

14. (a)

2

222

cm 169

cm 13

=

=AE

2

22222

cm 169

cm ]5)66[(

=

++=+ DEAD

∵ 222 DEADAE +=

∴ °=∠ 90ADE (converse of Pyth. theorem)

∴ AE is a diameter of the circle.


∴ The centre of the circumcircle lies on AE.

(b) ∵ CB ⊥ AD and AB = BD (given)

∴ CB is the perpendicular bisector of AD.

The centre lies on CB.

The centre also lies on AE.

∴ The intersection of CB and AE, i.e. C, is the

centre of the circumcircle.

15.

Join BD and DC.

∠ABD = 90° ∠ in semi-circle

∠ACD = 90° ∠ in semi-circle

∴ ∠ABD = ∠ACD

ACAB

ADAD

=

=

given

sidecommon

∴ △ABD ≅ △ACD RHS

∴ ∠BAD = ∠CAD corr. ∠s, ≅ △s

∴ AD bisects ∠BAC.

16. ∠ADC = p + q (ext. ∠ of △)

∠BCD = p (∠s in the same segment)

qp

pqp

BCDADCr

+=

++=

∠+∠=

2

(ext. ∠ of △)

17.

Let O be the centre of the circle and r cm be the radius.

Join OA.

cm rOA =

cm )1( −=

−=

r

MCOCOM

∵ ABOM ⊥ (given)

∴

cm 5

cm 102

1

=

×=

= MBAM

Consider △OAM.

13

262

1225

)1(5

22

222

222

=

=

+−+=

−+=

+=

r

r

rrr

rr

OMAMOA (Pyth. theorem)


18. (a)

°=

°−°−°=

∠−∠−°=∠

116

3628180

180 BACBCAABC ∠ sum of △

°=∠=∠ 116ADEABC

∴ A, B, C and D are concyclic. ext. ∠ = int. opp. ∠

∴ ABCD is a cyclic quadrilateral.

(b)

°=

°−°=

∠−∠=∠

62

54116

ACDADECAD (ext. ∠ of △)

°=

∠=∠

62

CADCBD (∠s in the same segment)

19. 1==∠

∠))

CD

DE

DAC

EBD arcs prop. to ∠s at ⊙ce

DACEBD ∠=∠

∴ P, A, B and Q are concyclic. converse of ∠s in the

same segment


54

(∠ at centre twice

∠ at ⊙ce)

20. Draw a line segment PQ in rectangle ABCD, then draw

another line segment RS which is perpendicular to PQ as

shown in the following figure.


Level 2

21.

°=

°−°=∠

°=∠+∠

40

140180

180

AOB

DOBAOB (adj. ∠s on st. line)

°=

∠=∠

40

AOBOBC (alt. ∠s, DA // CB)

°=

°×=

∠=∠

20

402

1

2

1AOBBCA

°=

°+°=

∠+∠=∠

60

2040

BCAOBCOKC (ext. ∠ of △)

°=

°−°=∠

°=∠+∠

120

60180

180

AKO

OKCAKO (adj. ∠s on st. line)

22. (a)

°=

°−°=∠

°=∠+∠

78

102180

180

AFC

AFCABC (opp. ∠s, cyclic quad.)

°=

∠=∠

78

AFCCDE (ext. ∠, cyclic quad.)

(b)

°=

°×=

∠=∠

156

782

2 CDECOE


ABOBAF ∠+∠

°=∠

°=°+°+°+∠

∠°×−=∠+∠+

64

36038156102

polygon) of sum (180)24(

BAF

BAF

FEOCOE

23. (a) °=∠ 90ADC (∠ in semi-circle)

°=∠

°=°+∠+°+°

°=∠+∠+∠

16

180)54(9020

180

BAC

BAC

BADADCAPD (∠ sum of △)

(b)

°=

∠=∠

16

BACBDC (∠s in the same segment)

°=

°−°=

∠−∠=∠

74

1690

BDCADCADB (ext. ∠ of △)

°=

°−°−°=∠

∠°=∠+∠+∠

52

5474180

) of sum (180

AKD

CADADBAKD △

24. (a) °=∠ 90ACB (∠ in semi-circle)

DCADBA ∠=∠ (∠s in the same segment)

°=∠+∠+∠ 180ECBCBECEB (∠ sum of △)

°=∠

°=∠

°=°+∠+°+∠+°

14

282

180)90()37(25

DCA

DCA

DCADBA

(b)

°=∠

°=°+°+∠

∠°=∠+∠+∠

53

1809037

) of sum (180

CFB

CFB

BCACBDCFB △

25.

Join MN.

∠ABM = ∠MNC (ext. ∠, cyclic quad.)

∠AEM = ∠MND (ext. ∠, cyclic quad.)

∠MNC + ∠MND = 180° (adj. ∠s on st. line)

°=

∠+∠=∠+∠

180

MNDMNCAEMABM

∴ ABMBAE ∠+∠

°=∠

°=∠+°+°

°×−=∠+∠+

115

36018065

180)24(

BME

BME

BMEAEM (∠ sum of polygon)

26.

Join BE.

))

CD

BC

CAD

BEC=

∠


°=

°=∠

42

)28(2

3BEC

))

CD

DE

CAD

DBE=

∠


°=

°=∠

56

)28(2

4DBE

∴

°=

°−°−°=∠

∠°=∠+∠+∠

82

5642180

) of sum (180

BKE

KBEKEBBKE △


55

(∠ at centre twice

∠ at ⊙ce)

27. (a)

21

1

+=

=∠

∠))

ABC

AB

AOC

AOB (arcs prop. to ∠s at centre)

°=

°×=

∠=∠

30

903

1

3

1AOCAOB

∴

°=

°×=

∠=∠

15

302

1

2

1AOBBCA

(b) ∵ OC = OA (radii)

∴ ∠ACO = ∠CAO (base ∠s, isos. △)

°=∠

°−°=∠

∠°=∠+∠+∠

45

901802

) of sum (180

CAO

CAO

AOCACOCAO △

∴

°=

°+°=

∠∠+∠=∠

75

3045

) of (ext. △AOBCAOCEO

28. (a) ∠APD = ∠CPB common angle

∠PAD = ∠PCB ext. ∠, cyclic quad.

∠PDA = ∠PBC ext. ∠, cyclic quad.

∴ △PAD ~ △PCB AAA

(b) ∠AKB = ∠DKC vert. opp. ∠s

∠BAK = ∠CDK ∠s in the same segment

∠ABK = ∠DCK ∠s in the same segment

∴ △AKB ~ △DKC AAA

(c) ∵ △PAD ~ △PCB (proved in (a))

∴DC

DC

ABPA

PD

DCPD

PA

PB

PD

PC

PA

+=

+=

+

+=

+

=

cm 8cm 12

cm 10)(6

cm 8

cm 8

cm 6

(corr. sides, ~ △s)

∴ cm 4=DC

∵ △AKB ~ △DKC (proved in (b)) ∴

cm 3cm 4

cm 10 BK

CK

BK

DC

AB

=

= (corr. sides, ~ △s)

∴ cm 5.7=BK

29. (a) ∵ AM = MB and CN = ND given

∴ ∠OMK = ∠ONK = 90° line joining

centre to mid-pt

of chord ⊥ chord

∵ AB = DC given

∴ OM = ON equal chords,

equidistant from

centre

OK = OK common side

∴ △OMK ≅ △ONK RHS

(b) ∴ KM = KN corr. sides, ≅△s

Also, BM = CN given

∴ KM – BM = KN – CN

∴ KB = KC

(c) KB = KC proved in (b)

Also, AB = DC given

∴ KA = KD

∴ ∠KAD = ∠KDA base ∠s, isos. △

∠BCD + ∠KAD = 180° opp. ∠s, cyclic quad.

∴ ∠BCD + ∠KDA = 180°

∴ BC // AD int. ∠s supp.

30. MDPNBP ∠=∠ ext. ∠, cyclic quad.

NMC

DMP

DPMMDP

NPBNBPBNP

∠=

∠=

∠−∠−°=

∠−∠−°=∠

180

180

s opp. vert.

of sum

of sum

∠

∠

∠ △△

∴ QNQM = sides opp. equal ∠s

31. (a)

Join OC.

112

11

++

+=

=∠

∠))

ABCD

ABC

AOD

AOC (arcs prop. to ∠s at centre)

∴

°=

°×=∠

45

904

2AOC

∵ OA = OC (radii)

∴ OCAOAC ∠=∠ (base ∠s, isos. △)

°=∠

°=°+∠

∠°=∠+∠+∠

5.67

180452

) of sum (180

OAC

OAC

AOCOCAOAC △

(b)

Join OB and AD.

∵ ))BCAB =

∴ ∠AOB = ∠BOC (equal arcs, equal ∠s)

∴

°=

°×=

∠=∠

5.22

452

1

2

1AOCAOB


56

°=

°×=

∠=∠

25.11

5.222

1

2

1AOBADB (∠ at centre twice ∠ at ⊙ce)

°=

°×=∠

=∠

∠

5.22

25.111

2CAD

AB

CD

ADB

CAD))


°=∠

°=°+°+∠

∠°=∠+∠+∠

25.146

18025.115.22

) of sum (180

AED

AED

EDAEADAED △

32.

Join BO and OE.

Let ∠CAE = x.

∠BOE = 2∠CAE ∠ at centre twice

= 2x ∠ at ⊙ce

∠ACE + ∠BOE = 180° opp. ∠s, cyclic quad.

∠ACE = x2180 −°

xCEA

CEAxx

CEACAEACE

=∠

°=∠++−°

°=∠+∠+∠

180)2180(

180 ∠ sum of △

∴ CEACAE ∠=∠

∴ CECA = sides opp. equal ∠s

33. (a) ∵ CDCE = given

∴

ABC

CDECED

∠=

∠=∠

quad. cyclic , ext.

isos. s, base

∠

∠ △

∴ ABAE = sides opp. equal ∠s

∴ △ABE is an isosceles

triangle.

(b) Let ∠ABD = x.

∵ 1==∠

∠))

AD

DC

ABD

DBC (arcs prop. to ∠s at ⊙ce)

∴

x

ABDDBC

=

∠=∠

x

DBCABDABE

2=

∠+∠=∠

x

ABEAEB

2=


x

AEBEDC

2=


)1(4 KKx

EDCAEBDCB

=

∠+∠=∠

(ext. ∠ of △)

°=∠ 90BDC (∠ in semi-circle)

)2(90

90180

) of sum (180

KKx

xDCB

BDCDCBDBC

−°=

°−−°=∠

∠°=∠+∠+∠ △

From (1) and (2), we have

°=°=∠

°=

−°=

72)18(4

18

904

DCB

x

xx

°=

°−°=∠

°=∠+∠

108

72180

180

BAD

DCBBAD (opp. ∠s, cyclic quad.)

34. ∠BDC = ∠ABD alt. ∠s, CF // BA

∠DEF = ∠ABD ext. ∠, cyclic quad.

∠KAE = ∠DEF corr. ∠s, CA // DE

∴ ∠BDC = ∠KAE

∴ A, K, D and F are concyclic. ext. ∠ = int. opp. ∠

35. (a) °=∠=∠ 90ADCABC ∠ in semi-circle

°=

∠−°=∠

90

180 ABCFBE adj. ∠s on st. line

°=

∠−°=∠

90

180 ADCEDF adj. ∠s on st. line

∴ EDFFBE ∠=∠

∴ BEFD is a cyclic

quadrilateral. converse of ∠s

in the same segment

(b) °=∠=∠ 35BACBDC (∠s in the same segment)

°=∠=∠ 35BDCBFE (∠s in the same segment)

°=

°+°=

∠+∠=∠

62

3527

CFECEFDCF (ext. ∠ of △)

°=∠

°=∠+°

∠=∠+∠

28

9062

CFD

CFD

ADCCFDDCF (ext. ∠ of △)

36. (a) ∵ ))DCAD = given

∴ DCAD = equal arcs, equal chords

∵ DADE = given

∴ DCDE =

∴ DCEDEC ∠=∠ base ∠s, isos. △

(b) ∵ AD = ED given

∴ DEADAE ∠=∠ base ∠s, isos. △

DCEDEC ∠=∠ proved in (a)

°=∠+∠+∠ 180CEBDECDEA adj. ∠s on st. line

DCEDAE

DECDEACEB

∠−∠−°=

∠−∠−°=∠

180

180

°=∠+∠ 180DAEBCD opp. ∠s, cyclic quad.

DCEDAEBCE

DAEDCEBCE

∠−∠−°=∠

°=∠+∠+∠

180

180)(

∴ BCECEB ∠=∠

∴ BC = BE sides opp. equal ∠s

37. (a)

°=

∠=∠

50

ABCADC opp. ∠s of // gram

°=∠+∠+∠ 180AECACECAE ∠ sum of △

°=

°−°−°=∠

130

1337180AEC


57

∵

°=

°+°=∠+∠

180

13050AECADC

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

(b)

Join DE.

°=

∠=∠

37

CAECDE (∠s in the same segment)

∵ DC = AB (opp. sides of // gram)

and AB = EC (given)

∴ DC = EC

∴

°=

∠=∠

37

CDEDEC (base ∠s, isos. △)

°=∠

°=°+°+∠

∠°=∠+∠+∠

106

1803737

) of sum (180

ECD

ECD

CEDCDEECD △

°=∠

°=°+°+∠

°=°+∠+∠

°=∠+∠

24

18050106

18050)(

180

ECB

ECB

ECDECB

ABCBCD (int. ∠s, CD // BA)

38. (a) (i) Consider △ABM and △CDM.

∠ABM = ∠CDM ∠s in the same segment

∠BAM = ∠DCM ∠s in the same segment

∠AMB = ∠CMD vert. opp. ∠s

∴ △ABM ~ △CDM AAA

∴ CM

AM

CD

AB= corr. sides, ~ △s

(ii) ∵ OL ⊥ AB given

∴ AL = LB

ABAL2

1=

∵ ON ⊥ CD given

∴ CN = ND

CDCN2

1=

Join LM and NM.

Consider △ALM and △CNM.

CM

AM

CD

AB

CD

AB

CN

AL

=

=

=

2

1

2

1

from (a) (i)

∠LAM = ∠NCM ∠s in the same segment

∴ △ALM ~ △CNM ratio of 2 sides, inc. ∠

(b)

Join OM, OR and OW.

∵ PM = MQ given

∴ OM ⊥ PQ line joining centre to

mid-pt. of chord ⊥ chord

∵

°=

°+°=

∠+∠

180

9090

RMORLO

∴ RLOM is a cyclic opp. ∠s supp.

quadrilateral.

∠ROM = ∠RLM ∠s in the same segment

∵

°=

°+°=

∠+∠

180

9090

WMOWNO

∴ WNOM is a cyclic

quadrilateral. opp. ∠s supp.

∠WOM = ∠WNM ∠s in the same segment

∵ △ALM ~ △CNM proved in (a)(ii)

∴ ∠ALM = ∠CNM corr. ∠s, ~ △s

∴ ∠ROM = ∠WOM

(c) Consider △ROM and △WOM.

WOMROM ∠=∠ proved in (b)

OMOM = common side

°=∠=∠ 90OMWOMR proved in (b)

∴ △ROM ≅ △WOM ASA

∴ RM = WM corr. sides, ≅ △s

∴ M is also the mid-point of RW.

Multiple Choice Questions (p. 2.73)

1. Answer: B

∵ OC ⊥ AB (given)

∴ AC = CB = 2 cm (line from centre ⊥ chord bisects

chord)

cm 12

cm 2422

22

=

−=

−= ACOAOC

(Pyth. theorem)

Area of △AOB

2

2

2

cm 34

cm 122

cm 12)22(2

1

2

1

=

=

×+×=

××= OCAB

2. Answer: B

For I, PQ ⊥ CD (given)

∴ I is true.

For II, ∵ AB = CD = 12 cm, OP ⊥ AB and OQ ⊥ CD

∴ OP = OQ (equal chords, equidistant from

centre)

∴ PQOQ2

1=

∴ II is true.

line from centre ⊥

chord bisects chord

line from centre ⊥

chord bisects chord


58

For III, ∵ OP ⊥ AB (given)

∴ AP = PB (line from centre ⊥ chord bisects

chord)

∴

cm 6

cm 122

1

2

1

=

×=

= ABAP

Consider △AOP.

cm 8

cm 610 22

22

=

−=

−= APOAOP

(Pyth. theorem)

cm 18

cm 16

cm 82

2

≠

=

×=

= OPPQ

∴ III is not true.

∴ Only I and II are true.

3. Answer: A

∠BCD = 90° (∠ in semi-circle)

°=

°−°=

∠−∠=∠

19

7190

ACBBCDDCA

°=

∠=∠

19

DCAABD


∴

°=

°+°=

∠+∠=

51

3219

BACABDx

(ext. ∠ of △)

4. Answer : D

For I, ))

CBDCCBDC :: ≠ in general

∴ I may not be true.

For II,

chords) equal arcs, (equal

2

1

2

1

CBAC

CB

CBCB

DCADAC

=

=

+=

+=

)

)))))

∴ AC : CB = 1 : 1

∴ II must be true.

For III, °=∠=∠ 90ACBADB (∠ in semi-circle)

∵ AC = CB (proved in II)

∴ ∠ABC = ∠CAB (proved in II)

∴ ∠ABC + ∠CAB

°=∠

°=°+∠

°=+∠

45

180902

180

CAB

CAB

ACB (∠ sum of △)

∴ 1:245:90: =°°=∠∠ CABADB

∴ III must be true. ∴ Only II and III must be true.

5. Answer: B

°=

°−°=∠

∠=∠+∠

19

2443EBD

BDCEBDBED

(ext. ∠ of △)

Join AD.

°=∠ 90ADB (∠ in semi-circle)

°=

°=+°+°+°

°=∠+∠

28

180)19()4390(

180

x

x

ABCADC


6. Answer: C For option C,

x

ADBACB

=

∠=∠


θ=

∠=∠ BACBDC (∠s in the same segment)

In △CDP,

zyx

xzxy

DCPPDCCPD

−−−°=

°=++++

∠°=∠+∠+∠

2180

180)()(

) of sum ( 180

θ

θ

△

∴ C must be true.

7. Answer: C

∠ADC + ∠ABC = 180° (opp. ∠s, cyclic quad.)

°=

°×=∠

∠−°=∠

+

+=

∠−°

∠

=∠

∠

∠−°=∠

75

18012

5

)180(7

5

86

37

180

180

ABC

ABCABC

ABC

ABC

ABC

ADC

ADC

ABC

ABCADC

))


8. Answer: A

Join BD.

°=∠ 90ADB (∠ in semi-circle)

°=

°×=

∠=∠

24

482

1

2

1CODCBD


°=

°+°=

∠+∠=

114

2490

CBDADBx

(ext. ∠ of △)


59

9. Answer: B

yDCB

BADDCB

−°=∠

°=∠+∠

180

180


∵ OD = OC (radii)

∴

y

OCDODC

−°=

∠=∠

180


x

ABODOC

=

∠=∠

(corr. ∠s, OD // BA)

∴

°−=

°=−°+−°+

°=∠+∠+∠

1802

180)180()180(

180

yx

yyx

OCDODCDOC (∠ sum of △) ∴ B must be true.

10. Answer: B

°=∠+∠ 180ABCADC (opp. ∠s, cyclic quad.)

yACD

AEDACD

xADC

−°=∠

°=∠+∠

−°=∠

180

180

180


°=+

°=°+−°+−°

°=∠+∠+∠

225

18045)180()180(

180

yx

xy

CADADCACD

11. Answer: B

°=∠=∠ 52CABCDB ∴ ABCD is a cyclic quadrilateral.

(converse of ∠s in the same segment)

°=∠

°=°+∠+°

°=∠+∠

56

18072)52(

180

ADB

ADB

CBAADC


12. Answer: D

For I, ∠ADE + ∠DEB = 180° (int. ∠s, AD // BE)

∠ABE = ∠DEB (given)

∴ ∠ADE + ∠ABE = 180°

∴ ABED is a cyclic quadrilateral. (opp. ∠s supp.)

For II, ∠ABE = ∠DEB (given)

∠DEB = ∠EFC (corr. ∠s, BE // CF)

∴ ∠ABE = ∠EFC

∴ BCFE is a cyclic quadrilateral.

(ext. ∠ = int. opp. ∠)

For III, ∠DFC = ∠ABE (ext. ∠, cyclic quad.)

∠ABE + ∠BAD = 180° (int. ∠s, AD // BE)

∴ ∠DFC + ∠BAD = 180°

∴ ACFD is a cyclic quadrilateral. (opp. ∠s supp.)

HKMO (p. 2.76)

1.

5

34 22

22

=

+=

+= BCABAC

(Pyth. theorem)

Consider △AMN and △ABC.

∠AMN = ∠NBC (ext. ∠, cyclic quad.)

∠ANM = ∠ACB (ext. ∠, cyclic quad.)

∠MAN = ∠BAC (common angle)

∴ △AMN ~ △ABC (AAA)

∴ AC

AN

BC

MN

AB

AM== (corr. sides, ~ △s)

Area of △AMN MNAM ××=2

1

Area of △ABC BCAB ××=2

1

∴ 2

4

1

4

1

2

12

1

of Area

of Area

=

×=

××

××=

AC

AN

BC

MN

AB

AM

BCAB

MNAM

ABC

AMN△△

∴

2

5

2

1

5

2

1

=

=

=

AN

AN

AC

AN

2

3

2

54

=

−=

−= ANABNB

2

53

2

45

94

9

32

3 2

2

22

=

=

+=

+

=

+= BCNBNC (Pyth. theorem)

∵ ∠NBC = 90° (given)

∴ NC is a diameter of the circle. (converse of

∠ in semi-circle)

∴ Radius of circle BNMC

4

53

2

53

2

1

2

1

=

×=

= NC


60

2. Let O be the centre of the circle.

Join OM, OA and OF. Let MN intersect OA at P.

Let BC = 2a.

Then AB = AC = BC = 2a,

and ∠B = ∠C = ∠BAC = 60° (prop. of equil. △)

∵ AM = MB = a and AN = NC = a (given)

∴ MN // BC and MN = BC2

1 = a (mid-pt. theorem) ∴

°=

∠=∠

60

BAMP

(corr. ∠s, MN // BC)

Since △ABC is an equilateral triangle, O is not only the

circumcentre of △ABC, but also its incentre, i.e. AO

bisects ∠BAC.

∴ °=°

=∠ 302

60MAP and AP ⊥ MN

In △AMP,

2

60cosa

AMMP =°=

and 2

360sin

aAMAP =°=

In △AMO,

3

32

3

2

30cos

aaAMAO ==

°= ∴ 6

3

2

3

3

32

a

aa

APAOPO

=

−=

−=

In △POF,

2

5

6

3

3

3222

22

a

aa

OPOFPF

=

−

=

−=

(Pyth. theorem)

∴ a

aa

PFMPMF

+=

+=

+=

2

51

2

5

2

2

512

51

+=

+

==a

a

MN

MFd

3. Let O be the centre of the circle.

Radius of circle

cm 5

cm 102

1

=

×=

Join OA, OB and OC.

Construct OD ⊥ BC and OE ⊥ AB.

OC = OB (radii)

CD = BD (line from centre ⊥ chord

bisects chord)

OD = OD (common side)

∴ △OCD ≅ △OBD (SSS)

∴ ∠COD = ∠BOD (corr. ∠s, ≅ △s)

OA = OB (radii)

AE = BE (line from centre ⊥ chord

bisects chord)

OE = OE (common side)

∴ △OAE ≅ △OBE (SSS)

∴ ∠AOE = ∠BOE (corr. ∠s, ≅△s)

Let ∠COD = α and ∠AOE = β.

Then, CD = 5 sin α and AE = 5 sin β

)sin(sin10

sin10sin10

22

βα

βα

+=

+=

+=+ CDAEBCAB

AB + BC attains its maximum when α = β = 45°.

∴

cm 210

cm 2

2

2

210

)45sin45(sin10

=

+=

°+°≤x

∴ The greatest possible value of x is 210 cm.

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