Chapter 3

- parts of a circle.

3.1 properties of circles.- area of a sector of a circle.

the area of the smaller sector can be found by the following formula:

†

A =q º

360ºpr2 , given

†

q in degrees, or!

†

A =q

2ppr2 , given

†

q in radians

- the Chord Right Bisector Property (RBP).

†

OM^AB iff

†

AM = MBie. the centre of a circle always lies on the perpendicular bisector of any chord.

proof:I. given

†

OM^AB

because

†

DOAM ≡ DOBM (HS

†

≡ ),therefore,

†

AM = MB (

†

≡ Ds)

II. given

†

AM = MB

because

†

DOAM ≡ DOBM (SSS),

†

q = j (

†

≡ Ds)therefore,

†

q = j = 90º

ex. prove that

†

XY is the right bisector of

†

PQ .

†

DXPY ≡ DXQY (SSS)

therefore

†

–PXM = –QXM = q

so

†

DXPM ≡ DXQM (SAS

†

≡ )

ergo,

†

–XMP = –XMQ = 90º and

†

PM = MQ (

†

≡ Ds)

QED

- the Equal Chords Property (ECP).

†

AB = CD iff

†

d1 = d2

proof:I. given

†

AB = CD

†

ND = MB

†

DOBM ≡ DODN

†

OM = ON (

†

≡ Ds)

II. given

†

OM = ON

†

DOMB ≡ DONC (HS

†

≡ )

therefore,

†

MB = NCsimilarly,

†

AM = DN

thus

†

AM + MB = DN + NCAB = DC

3.2 angles in a circle.

- some definitions.

†

q is called the central angle, and it is said to be subtended by minor arc

†

AB .

the reflex angle

†

–AOB is the central angle subtended by major arc

†

AB .

†

–AXB is an inscribed angle subtended by arc

†

AYB .

†

–AYB is an inscribed angle subtended by arc

†

AXB .

- Angles in a Circle Properties (ACP).

1. central angles are twice the size of inscribed angles subtended by the same arc.

2. inscribed angles subtended by the same arc are equal.

3. inscribed angles subtended by a diameter equal 90º.

proof:because

†

OC = OA = OB ,

†

DOAC and

†

DOCB are isosceles

thus,

†

–ACO = –CAO and

†

–OBC = –OCBby EAT,

†

–DOA = 2 –OAC( ) and

†

–DOB = 2 –OCB( )

ergo,

†

–AOB = 2 –OAC( ) + 2 –OCB( )= 2 –ACB( )

QED.

ex. find the value of

†

w ,

†

x ,

†

y ,

†

z .

answers:

†

w = 40º ,

†

x = 70º ,

†

y = 60º,

†

z = 30º

ex. prove

†

DACE is isosceles.in

†

DEBD ,

†

ED = BDtherefore

†

d = e (ITT)

but

†

g = d , ergo

†

g = d = e

†

a = e because

†

a and

†

e are subtended by the same arc

therefore

†

a = g by transitivity, hence

†

DACE is isosceles. QED.

ex. prove that

†

AB = BC .

because

†

OA and

†

OC are radii of a common circle,

†

OA = OC .

†

OB is common to both

†

DOBC and

†

DOBA .

because

†

OA is a diameter, by ACP,

†

–OBA = 90ºbecause

†

OB^AC ,

†

\ AB = BC (RBP).

ex. determine the size of the central angle subtended by an arc of

†

2p if the circle’s radius is 6.

†

C = pd=12p

, therefore

†

q360º

=2p

12pq

360º=

16

q = 60º

3.3 cyclic quadrilaterals.

- concyclic points are points which lie on the circumference of the same circle.

- cyclic polygons have concyclic vertices.

ex. prove that all triangles are cyclic given any

†

DABCie. prove

†

OA = OB = OCclearly,

†

OA = OB (RBT)

similarly,

†

OB = OC (RBT)

by transitivity, therefore

†

OA = OB = OC QED.

ex. prove that if a quadrilateral

†

ABCD is cyclic, then its opposite angles are supplementary.

†

2q + 2b = 360ºq + b =180º

QED.

- Cyclic Quadrilateral Properties.

1. opposite angles are supplementary

2. any exterior angle is equal to the opposite interior angle

3. a side subtends equal angles at the remaining vertices

ie.

†

–BAC = –BDC or

†

–CAD = –CBD

ex. in a cyclic quadrilateral

†

ABCD ,

†

AB = AD ,

†

–BCD =110º ,

†

–BAC = 30º . find

†

–ABC .

†

a = 40º (CQP) and

†

j = 55ºbut

†

CD subtends both

†

–CAD and

†

–CBDso

†

–CAD = –CBD = 40ºtherefore,

†

–ABC = 55º+40º+95º

ex. show that

†

BCED is cyclic.

in

†

DABE ,

†

–A = 70ºin

†

DABC ,

†

50º+70º+2q =180ºq = 30º

†

–BEC =100º= –BDCbecause side

†

BC subtends equal angles at

†

D and

†

E , therefore

†

BCED is cyclic.

3.5 tangent properties.

- Tangent/Radius Properties (TRP)

†

AB is a tangent iff

†

OP^AB

proof:I. given

†

AB is a tangentassume

†

–OPA ≠ 90ºWLOG let

†

–OPA < 90ºconstruct

†

Q on

†

AB such that

†

–OQP = –OPAergo,

†

OQ = OP (ITT)

†

ƨ

therefore

†

OP^AB

II. given

†

OP^BAassume

†

AB is a secant.

clearly,

†

OQ = OPso

†

–OQP = –OPQ = 90º

†

ƨ

therefore,

†

AB is a tangent.

QED.

- Tangent from a Point to a circle Property (TPP)

IF

†

PA and

†

PB are tangent segments,

THEN

†

PA = PB

proof:in

†

DAOP ,

†

DBOP

†

AO = BO (radii of a common circle)

†

OP is common and

†

–OAP = –OBP = 90º (TRP)

therefore,

†

DAOP ≡ DBOP (HS)

whence

†

AP = BP , QED.

ex. if

†

PA =15, determine the perimeter of

†

DPCD .clearly,

†

DE = DB and

†

EC = CA (TPP)

the perimeter of

†

DPCD

†

= PD + DE + PC= PD + DE + EC + PC= PD + DB( ) + CA + PC( )= PB + PA= 30

- Tangent Chord Property (TCP)

the angle formed by a chord and a tangent is equal to the inscribed angle

subtended by the chord in the opposite segment.

ie.

†

–BAD = –BXA–BAC = –BYA

proof:

clearly,

†

a = b (ACP), and

†

–YBA = 90º (ACP)

†

b + g = 90º (SATT)

but

†

g + d = 90º (TRP)

whence

†

d = b = a . QED.

- Intersecting Chords Property (ICP)

IF two chords

†

AB and

†

CD intersect at

†

ETHEN

†

AE ¥ EB = CE ¥ ED

proof:

†

–AEC = –DEB (opposite angles) and

†

–ACD = –ABD (subtended by common arc)therefore

†

DACE ~ DDBE (AA

†

~ ).

thus

†

AEED

=CEEB

(

†

~ D s)

whence

†

AE ¥ EB = CE ¥ ED (POP) QED.

- Intersecting Secants Property (ISP)

IF two secants

†

AB and

†

CD meet at

†

PTHEN

†

AP ¥ PB = CP ¥ PD

- Corollary to ISP

†

AP ¥ PB = PT( )2

ex. prove that

†

ADQP is cyclic.clearly,

†

DXBC ~ DXDA and

†

a = g

since

†

XQQB

=XPPC

=11

therefore

†

QP || BC (DST)so

†

b = g (PLT, corresponding angles)thus

†

a = bwhence

†

ADQP is cyclic because

†

a and

†

b are subtended by side

†

AP

for review questions, see:p. 112 #1-7p. 108 #1, 2, 4, 5, 7, 9-12, 14-16, 18, 21