Circle - Holy Cross School resources/Mathematics Conten… · Line through a point on a circle is a tangent to the circle line is perpendic ular to the radius. x Theorem 3 and its

Grade 11 CAPS

Mathematics

Video Series

Circle

Geometry II

In this Video we will :

Lessons linked to this Video

and theorems that are

linked to

and related .

Investigate prove

cyclic quadrilaterals

solve ride Lessrs on 1

and theorems that are

linked to and

related .

Investigate prove

tangents of circles

solve riders Lesson 2

Grade11 CAPS

Mathematics

Video Series

Lesson 1

Theorems

On

Cyclic

Quadrilaterals

In this lesson we will :

Outcomes for Lesson 1

Recap the linked quadrilaterals and cyclic quadrilaterals. terminology

Solve riders related to Theorems 1, 2 and its converses.

Prove that :

,

(

If the sum of the two opposite angles of a

quadrilateral is equal to 180 its vertices are concyclic.

Converse o f Theorem 1)

Investigate and prove that :

(

The exterior angle of a cyclic

quadrilateral is equal to opposite interior T ang heorle. em 2)

Prove that :

(

If the exterior angle of a quadrilateral is equal

to its to opposite interior angle, the quadrilateral is cyclic.

Converse of Theorem 2)

Investigate and prove that :

(

The sum of the opposite angles

of a cyclic quadrilateral is equal t Theoro 180 . em 1)

Any four sided polygon is a .quadrilateral

Quadrilateral Terminology (Undefined terms)

1, 2, 3 and 4 are the

of quadrilateral .

1 2 3 4 360

ABCD

interior angles

Know that :

5, 6, 7 and 8 are the exterior angles

of quadrilateral .

5 6 7 8 360

ABCD

Know that :

Points lying on a circle are .

A quadrilateral having its vertices on a circle

is called a .

concyclic

cyclic quadrilateral

Inscribed angles on the same side of a chord are equal.

Revisit Theorem 5 (Video Lesson : Circle Geometry 1)

Quadrilateral with , , and are concyclic or

Quadrilateral with is a cyclic quadrilateral or

If

ABCD ADB ACB A B C D

ABCD ADB ACB ABCD

Consider version 4 of Theorem 5 in formulation of converse :

the line segment joining two points subtends equal angles at two other points

on the same side of it, the four points must be concyclic.

:Converse of Theorem 5 (Proved in next slide)

1 Inscribed angles on the same side of a chord are equal (original).

2 Inscibed angles on a chord in the same segment are equal.

3 , , and concyclic .

4 a

A B C D ADB ACB

ABCD

Theorem 5 can be reformulated :

cyclic quadrilateral .ADB ACB

If line segment joining two points subtends equal angles

at two other points on the same side of it, the four points must be concyclic.

Converse Theorem :

s subtending equal and at

and on the same side of .

, , , and are a concyclic.

Draw and assume that .

BC BDC BAC

D A BC

A B C D

DBC A DBC

Given :

Aim is to prove that :

Construction :

Let cut at and draw . DBC BA M MC

:

1 inscribed

Assume that and

s n

1

o

M DBC A DBC

BDC BC

Proof

A quadrilateral has is a cyclic quadrilateralABCD BAC BDC ABCD

Combining Theorem 5 and its Converse we have :

Ext. 1 of int. opp. AMC A

This is a contradiction because 1 of 1AMC A ACM A

Assumption 1 is incorrect and thus , , and are concyclic.A B C D

Sum of opposite angles of a cyclic

quadrilateral is equal to 180 (or supplementary).

Investigation :

for more

such investigations!

GeoGebra

Suggested Conclusion :

180Cyclic quadrilateral

180

A CABCD

B D

Conjecture 1 :

Any with cyclic quadrilateral .

180

180

Draw and

O ABCD

A C

B D

DO BO

Given :


Construction :

:

1 2 central 2 inscribed C

Proof

s2 2 2 1 360 round a pointA C

sBut 360 of a quadrilateral A C B D

Sum of opposite angles of a cyclic

quadrilateral is equal to 180 (or supplementary).

Theorem :

180Cyclic quadrilateral

180

A CABCD

B D

Theorem 1 :

2 2 central 2 inscribed A

180A C

180 180 B D A C

Assumption in 1 was incorrect must fall on passes throug M D ABC D

If the sum of the two opposite angles of a

quadrilateral is equal to 180 its vertices are concyclic.

Theorem :

but 180 givenD B

1 which is impossible 1 1D D MCD D

Any quadrilateral having 180 .

is a cyclic quadrilateral.

Draw and assume that .

Let cut at and dra

ABCD B D

ABCD

ABC D ABC

ABC AD M

Given :


Construction :

w . MC

Assume : that and 1M ABC D ABC Proof

1 180 , , and are concyclicB A B C M

Quadrilateral with opposite angles supplementary Quadrilateral is cyclic

Converse of Theorem 1 :

Quadrilateral is cyclic sum of opposite angles of quadrilateral is 180


The exterior angle of a cyclic quadrilateral

is equal to the interior opposite angle.

Investigation :

From investigations we make the :

Exterior angle of a cyclic quadrilateral

be equal to opposite interior angle.

conjecture

seems to

for more

such investigations.

GeoGebra

Cyclic quadrilateral and

with .

Exterior

interior opposite

ABCD

AE B AE

CBE

CDA

Given :


2 180 straight CBE ABE

Quadrilateral cyclic exterior interior opposite

Theorem 2 :

The exterior angle of a cyclic quadrilateral

is equal to the interior opposite angle.

Theorem :

s

:

1 2 180 sum of int. opp. of cyclic quad.

Proof

1 2 2 1CBE CDA CBE

Quadrilateral having drawn

with and .

is a cyclic quadrilateral.

ABCD DC

E DC BCE DAB

ABCD

Given :


:

1 180 straight BCE DCE

Proof

If the exterior angle of a quadrilateral is equal

to its interior opposite angle, the quadrilateral is cyclic.

Theorem :

1 180 given:

or 2 1 180

DAB BCE DAB

sQuadrilateral is cyclic sum of opp. 180ABCD

A quadrilateral is cyclic Exterior angle is equal to opposite interior angle


exterior interior opposite Quadrilateral cyclic


s

1

on chord

or similar results linked to chords , and

BAC BDC BC

AB AD DC

ABCD is a cyclic quadrilateral if we can prove that :

Sufficient conditions for a quadrilateral to be cyclic

2

180 Opposite angles supplementary

or 180

A C

B D


3

Ext. Opp. int.

or similar results linked to , , and

their respective opp. ext. angles.

DCE A

B C D


Riders mainly linked to Theorem 1 and its converse

Quadrilateral is cyclic sum of opposite angles of quadrilateral is 180

Theorem 1 and its Converse :

s

180 60 70 50

Sum of in

e

s

180 180 65 115

Opp. of cyclic quad.

b a

s

5 3 180

Opp. of cyclic quad.

c d

36

and 60

c

d

s

70 180 110

opp. of cyclic quad.

f f

s

60 35 180


g e

180 60 50 35

35

g

13065

2

1 at circum at centre

2

a

Find the value of each letter in the riders.

Rider 1 Rider 2Rider 3

A quadrilateral is cyclic Exterior angle is equal to opposite interior angle

Theorem 2 and its Converse :

Riders linked to mainly Theorem 2 and its converse

113

ext. opp. int.

f

s

180 95

co-int.

b a

88

ext. opp. int.

d

s

180 67 113


e

85

ext. opp. int.

a

Find the value of each letter in the riders.

s

180 85


c b

92

ext. opp. int.

g

Rider 1

Rider 2 Rider 3

Tutorial 1: Find the value of the letters in each of the following riders

PAUSE Video

• Do Tutorial 1

• Then View

Solutions

Tutorial 1: Riders 1 and 2: Suggested Solutions

Determine the values of the angles represented by the letters.

s6 180 Opp. of cyclic quad.

30

a

a

2 at centre 2 at circum

60

b a

b

s4 180 Opp. of cyclic quad.

45

c

c

s4 80 180 Opp. of cyclic quad.

4 100

25

d

d

d

Tutorial 1: Riders 3 and 4: Suggested Solutions


s 1 180 70 Opp. of cyclic quad.

1 110

s 2 180 110 70 sum of in e

sBut 2 35 base of isosceles e

1 ext. opp. int. f

sBut 1 85 180 co-interior

1 180 85 95

1 95f

Tutorial 1: Rider 5: Suggested Solution


88 ext. opp. int. g

s180 co-interior

180 180 88 92

g h

h g

s180 Opp. of cyclic quad.

180 92 88

i h

i

iCan you see an alternative method to find ?



s 1 2 180 co-interior

180 1 2 180 72 90 18

k

k

s

2 90 18 108

3 180 2 Opp. of cyclic quad.

k

j k

1 72 ext. opp. int.

2 90 in semi-circle

s 3 alternate k

Label angles

180 108 18 3 18 54j k j

Grade11 CAPS

Mathematics

Video Series

Lesson 2

Theorems

On

Tangents

In this lesson we will :

Outcomes for Lesson 2

Recap the and linked to tangents. terminology axioms

Investigate and prove ( :

Line through a point on a circle is a tangent to the circle line is perpendicular to the radius.

Theorem 3 and its converse)

⇔

Solve riders related to Theorems 3, 4 and 5 and its converses.

Prove the following linked to theorem 3 and its converse :

1 Tangents drawn from an external point to a circle are equal.

2 The bisector of the angle between the two tangents passes

corollaries

through the centre of the circle.

3 The line segment joining the centres of two circles cutting each other is the perpendicular

bisector of the common chord.


The bisectors of the interior angles of a triangle are concurrent.

Theorem 4)

Investigate and discuss circle and circles of a given triangle. inscribed escribed


A line is a tangent to a circle

Angle formed between a line, that is drawn through the end point of a chord and the chord,

is equal to the an

Theorem 5 and its converse)

gle subtended by the chord in the alternate segment.

Terminology linked to tangents

is a as it cuts in two points and . AB O A Bsecant line

If approaches until it may be considered

to coincide with , the secant is a to

the at the point of coincidence.

B A

A tangent

The point of coincidence is called the

.point of contact

We may say:

A tangent is a line (line segment) which

has only one point in common with the .

Shortest distance from a point to a line

Axiom:

1 The is the shortest

line segment from a point to a line.

2 Conversely, the shortest line segment

from a point to a line is the perpendicular.

perpendicular


If a line through a point on a is a tangent to the

then the line is perpendicular to the radius.

conjecture

for more


GeoGebra

If a line through a point on a is a tangent

to the , then the line is perpendicular to the radius.

Investigation :

If line through a point on a is a tangent to the

then the line is perpendicular to the radius.

Theorem 3 :

If a line through a point on a is a tangent

to the then the line is perpendicular to the radius.

Theorem :

is a tangent with the point of contact.

.

Draw with

AB O P

AB OP

OT T AB

Given :


Construction :

:

lies outside is contact pointT O P

Proof

is the radius of OP OT OP O

is the shortest of all segments drawn from to .OP O AB

Perpendicular is shortest distance between a point and lineOP AB

A line drawn through any point of a

circle perpendicular to the radius of the is a tangent to the .


A line drawn through any point of a circle

perpendicular to the radius of the is a tangent to the .

Theorem :

and any point on this circle.

with .

is a tangent to at .

Draw with

O P

P AB OP AB

AB O P

OT T AB

Given :


Construction :

: given

is shortest distance from to

AB OP

OT OP OP O AB

Proof

radius of lies outside OT O T O

Any point on , except , lies outside is a tangent to AB P O AB O

Line through a point on a circle is a tangent to the circle line is perpendicular to the radius.

Combining Theorem 3 and its converse :

⇔

linked to Theorem 3 and its converse.Corollories

Tangents drawn from an external point to a circle are equal.Corollory 1:

The bisector of the angle between the two tangents

passes through the centre of the circle.

Corollory 2:

The line segment joining the centres of two circles cutting each

other is the perpendicular bisector of the common chord.

Corollary 3 :

and tangents to AP BP O

AP BP

bisectsPC APB

O PC

and with common chord

and

O P AB

OP AB AC BC

Proof of Corollary 1.

Tangents drawn from an external point to a circle are equal.Corollory 1:

and tangents to AP BP O

AP BP

Given :


sIn and ,OAP OBPProof :

radii

common

tangent radius

AO BO

OP OP

OAP OBP

90 , ,AOP BOP s s AP BP

and tangents to AP BP O AP BP


The bisector of the angle between the two tangents

passes through the centre of the circle.

Corollory 2:

bisectsPC APB

O PC

Given:

Aim is to prove that:

Thus we assume that

is not a

Assume that

bisector of .

PO

APB

O PCProof :

radii

common

tangent radius

AO BO

OAP OBP OP OP

OAP OBP

APO BPO

and tangents to and bisect AP BP O PC APB O PC

also bisectsPO APB O PC


The line segment joining the centres of two circles cutting each

other is the perpendicular bisector of the common chord.

Corollary 3 :


and

O P AB

OP AB AC BC

Given:

Aim to prove that:

sIn and ,ACP BCPProof :

proved

common

proved

AP BP

PC PC

APC BPC

, ,ACP BCP s s

90ACP BCP

AC BC


and

O P AB

AB OP AC BC


The bisectors of the three interior angles of any triangle

are concurrent ( or intersect in a common point).

conjecture

for more such investigations.GeoGebra

The bisectors of the interior angles of a triangle are concurrent.

Investigation :

The bisectors of the interior angles of a triangle are concurrent.Theorem 4 :

The bisectors of the interior angles of

a triangle are concurrent.

Theorem :

Any .

Bisectors of the angles are concurrent.

Draw and the bisectors of and meeting in .

Draw and perpendiculars , and to , and

ABC

BO AO B A O

OC OD OE OF AB AC

Given :


Construction :

respectively.BC

:

, ,BDO BFO s OD OF

Proof

, ,ADO AEO s OD OE

OF OD OE

In and :CEO CFO

90 construction

common

proved

CEO CFO

OC OC

OE OF

, ,90CEO CFO s s

is a angle bisector of meeting other two angle bisectors in . ECO FCO CO C O

Bisect any two angles and

determine the in-centre .O

circle of a gi .ven triangleConstruction : Inscribed

Drop a perpendicular to any of

the three sides to determine the

radius of the inscribed .r OD

With as centre and

as radius draw inscribed .

O OD , and are tangents

AB AC BC

r OE OD OF

Bisect any two of indicated angles to determine the centre.

Drop perpendicular to determine the radius .

Draw escribed circle (Take note of three tangents).

r OF

circles of a given triangle.Construction : Escribed

Sketches of other two left as an exercise.

Take note of tangents in each case.

Three possibilities.

Example 1: Lengths of tangents and inscribed circles.

The inscribed touches , and of at , and respectively.O AB BC AC ABC D F E

1 If 8 units, 12 units and 9 units

determine the lengths of , and .

AB BC AC

AD BF CF

, and

Tangents from exterior points are equal

AD AE CE CF BD BF

8 8 1

9 9 2

and 12 3

AD BD AD BF

AE EC AD CF

CF BF

2 1 1 4

3 4 2 13 6.5

CF BF

CF CF

Back substitution

12 12 6.5 5.5

8 8 5.5 2.5

BF CF

AD BF

Check correctness :

2.5 5.5 8

2.5 6.5 9

5.5 6.5 12

AB AD DB

AC AE EC

BC BF FC

Example 2 : Lengths of tangents and inscribed circles.

The inscribed touches , and of at , and respectively.

Assume that , and .

O AB BC AC ABC D F E

AB c AC b BC a

2 If prove that .2

a b cS AD S a

, and


AD AE CE CF BD BF

Perimeter of Perimeter of 2

2 2

ABC a b cS ABC S

2

2 2 2 2 ; ;

S AD AE BD BF CE CF

S AD BF CF AD AE BD BF CE CF

S AD BF CF

S AD BF CF AD BF CF AD a AD S a

Example 3 : Lengths of tangents and inscribed circles.

The inscribed touches , and of at , and respectively.

Assume that , and .

O AB BC AC ABC D F E

AB c AC b BC a

area of

areas of

ABC

BOC AOC AOB

3 If prove that area of .2

a b cS ABC rS

2 2 2

ar br cr

2

a b cr rS

Tutorial 2: Tangents, Inscribed and Escribed circles

1 The sides of a quadrilateral are tangents to .

Prove that .

ABCD O

AB CD AD BC

2 is the centre of the inscribed circle of .

The circle touches , and at , and

respectively. , and are drawn.

2.1 Make a rough sketch.

2.2 Prove that

O ABC

AB BC CA P Q R

OP OQ OR

.2

B CPQR

3 is an escribed circle of with radius .

If 2 , prove that:

3.1 .

3.2 and

3.3 area of .

DEF ABC r

S a b c

AD AF S

CE S b

ABC r S a

PAUSE Video

• Do Tutorial 2

• Then View Solutions

Tutorial 2: Problem 1: Suggested Solution

1 The sides of a quadrilateral are tangents to .

Prove that .

ABCD O

AB CD AD BC

AB CD

AE EB DG CG

from known results - see figureAH BF HD CF

commutative & associative

properties for additionAH HD BF CF

AD BC

: Proof

2 2

PQR PQO OQR

B C

Tutorial 2: Problem 2: Suggested Solution

2 is the centre of the inscribed circle of .

The circle touches , and at , and

respectively. , and are drawn.

2.1 Make a rough sketch.

2.2 Prove that

O ABC

AB BC CA P Q R

OP OQ OR

.2

B CPQR

s180 Opp. of cyclic quad.POQ B

s180 Sum of in PQO QPO POQ OPQ

2 180 180 2

BPQO B OP OQ PQO

Similarily because is a cyclic quad. 2

CROQC OQR

sis a cyclic quadrilateral Opp. supplementaryPOQB

Tutorial 2: Problem 3.1: Suggested Solution

3.1 is an escribed circle of with radius .

If 2 , prove that .

DEF ABC r

S a b c AD AF S

, and


AD AF BD BE CE CF

2 2AD AF AD AF AD AF

2 2AD AF AB BD AC CF

2 2 ;AD AF AB BE AC EC BD BE CF EC

2 2AD AF AB AC BE EC

AB AC BC c b a

2

a b cAD AF S



If 2 , prove that .

DEF ABC r

S a b c CE S b

AF AC CF

AC EC CF EC

EC AF AC

Proved in 3.1 that .AF S

EC AF AC

S b AC b

: Proof



If 2 , prove that area of .

DEF ABC r

S a b c ABC r S a

area area and

area area and

area area and

BDO BEO BD BE h r

CEO CFO EC CF h r

DAO FAO AD AF h r

area

2 area 2 area 2 area

ABC

ADO BDO CEO

AD r BD r CF r

r AD BD CF

r AB BD BD CF

r AB CE

2

2 1

S a b c

b c S a

...

1 : 2r S a S r S a From ( ; )r c S b CE CF AF S


If a line is a tangent to a circle then the angle between a chord

and the tangent, drawn at the point of contact of the chord, is equal

to the inscribed angle

conjecture

which the chord subtends in the alternate segment.

for more


GeoGebra

The angle between a tangent and a chord, drawn

at the point of contact of the chord, is equal to the inscribed

angle which the chord subtends in the alternative segment.

Investigate :

Angle between tangent and chord is equal to the inscribed angles

which the chord subtends in alternate segment.

Theorem 5 :

Angle between tangent and chord is equal to the inscribed

angles which the chord subtends in alternate segment.

Theorem :

Any with chord and

tangent with point of contact .

and

.

Draw diameter and .

O BD

AC B

DBC F

ABD M

BE DE

Given :


Construction :

:

90 angle in a semicircle

1 90 sum s in a

EDB

E

Proof

180 180ABD DBC ABC

1 90

tangent and radius

DBC

BC OB

inscribed s on chord DBC E F BD

180 is a cyclic quad.F M FBMD

180

180

ABD DBC

F DBC F

M

If the angle formed between a line, that is drawn

through the endpoint of a chord, and the chord, is equal to the angle subtended

by the chord in the alternate segment, then the line

Converse Theorem :

is a tangent to the circle.

and chord such that

in alternate arc.

is a tangent to at .

Draw a tangent at .

ABC AB

TAB ACB

AT ABC A

AE A

Given :


Construction :

:

angle between tangent and chord

inscribed in alternate arc

AE ABACB EAB

Proof

But GivenACB TAB

EAB TAB

This is only possible if and coincide.AE AT

is a tangent to at .AT ABC A

If the angle formed between a line, that is drawn

through the endpoint of a chord, and the chord, is equal to the angle subtended

by the chord in the alternate segment, then the


line is a tangent to the circle.

Sufficient condition for a line (segment) to a circle to be a tangent.

1 If

then is a tangent to .

BA OA

BA O

2 If or

then is a tangent to .

CBD A ABE C

ED ABC

Riders linked to Tangents

9018

5a

90

angle in semi-circle

d

40

angle between tangent & chord

equal to angle in opp. segment

b

80

isosceles

g f

180

sum 180

180 160 20

e g f

e

5 90

radius tangent

sum 180

a

Find the value of the angles indicated by each letter in the riders.

Where applicable is the centre of the circle and a tangent.O AB

90 40 50

radius tangent or sum 180

c

80

angle between tangent & chord

equal to angle in opp. segment

f

More Riders linked to Tangents

70

1 at circum. at centre

2

q

180 2 48 84

angles in isosceles

j

30

angles on same chord

l

90 60

angle in semi-circle

and sum 180

m l

48

tangent & chord

in opp. segment

i



84

tangent & chord

in opp. segment

h j

30

chord & tangent

in opp. segment

k

70

chord & tangent

in opp. segment

n q

90 20

alt. s and radius tangent

p q

Tutorial 3: Riders linked to Tangents

PAUSE Video

• Do Tutorial 3

• Then View Solutions




Find the value of the angles indicated by each letter in the rider.


65

tangent & chord in opp. segment

a

55


b




38


Note: and angles on same chord

d e

d e

90 38 52

radius tangent

c




1

2

1 4


f

2 alt. 1 4

parallel line segments

f

2 10 4 180

straight angle

f f 4 10 4 180

18 180

10

f f f

f

f




1

50


h

1 2 100

at centre 2 at circum.

h

180 1

2

isosceles

sum 180

g

180 10040

2g




40


k

40 60 100


j

180

opp. s in cyclic quad. supplementary

i j

180 100 80i

End of the Second Video on Circle Geometry

REMEMBER!

•Consult text-books for additional examples.

•Attempt as many as possible other similar examples

on your own.

•Compare your methods with those that were

discussed in this Video.

•Repeat this procedure until you are confident.

•Do not forget:

Practice makes perfect!

Documents

Circle - Holy Cross School resources/Mathematics Conten… · Line through a point on a circle is a tangent to the circle line is perpendic ular to the radius. x Theorem 3 and its