Circle Geometry I Grade 11 CAPS Mathematics Series Outcome for this Topic

In this Topic we will :

• Investigate and prove theorems linked to perpendicular bisectors of chords and solve related riders . Unit 1

• Investigate and prove theorems linked to inscribed and central angles and solve related riders . Unit 2 Unit 1

Theorems

on Grade 11 Perpendicular CAPS Bisectors of Mathematics Series Chords Outcomes for Unit 1 In this Unit session we will : • Discuss the features of Geometric Axiomatic System . • Recap the terminology linked to circles. • Investigate and prove that : Segment from centre of circle perpendicular to chord bisects chord. (Theorem 1) • Investigate and prove that : Segment from centre of a circle to midpoint of chord is perpendicular to the chord. (Theorem 2) • Investigate and prove that : Perpendicular bisector of chord passes through the centre of a circle. (Theorem 3) • Solve riders related to Theorems 1, 2 and 3. Features of a Geometric Axiomatic System An Axiomatic System consists of some: Undefined terms Defined terms Axioms (Accepted unproved statements or proved theorems) Theorems (proved mathematical statements) Examples from Geometry : Undefined terms :Point, line (segment; ray), angle, triangle, exterior and interior angles of triangle Defined terms : A straight angle measures 180

Axiom 1: 12 corresponding angles are equal Axioms :  Axiom 2: 12 alternate angles are equal Theorem 1: The sum of the interior angles of a triangle   is equal to 180 Theorems :  Theorem 2: The exterior angle of a triangle is equal to the  sum of the two interior opposite angles Prove that : Given ABC then  A   B   C  180 

Given : ABC Aim is to prove that : ABC      180  Draw BD with C BD Construction :  Draw CE BA Proof: Note : A proof is a sequence of logical steps ABC     A sound reason needs to be given  A   B   ACB  To avoid confusion for each of the logical steps  ACE   B   ACB  Axiom 2:  A  alternate  ACE  ACE   DCE   ACB  Axiom 1:  B  corresponding  DCE  BCD 180   Definition:  BCD is a straight angle Once proved this theorem can be accepted as an additional axiom. Proof of the second theorem, namely that ACD   A   B, is left as an exercise. Circle Terminology (Undefined terms)

Given a circle with centre O  Refer to this as O The radius , r , is the distance from the centre of the circle to any point on the circumference. An arc is a part of the circumference of a circle.

A chord is the line segment joining the ends of an arc. Note: Chord can be linked to two arcs. Or a chord is a line segment connecting two points on .

The diameter is a special chord that passes through the centre of the circle. A segment is the part of the circle that is cut off by a chord. A chord divides the circle into two segments. Investigation : Segment from centre of a circle and perpendicular to a chord bisects the chord.

Suggested Conclusion : Many more investigations possible by means of  GeoGebra.

Conjecture 1: OMchord AB  AM  BM Theorem : Segment from centre of a circle and perpendicular to a chord bisects the chord.

Given : Any O with OM chord AB Aim is to prove that : AM BM Construction : Draw OA and BO Proof: AM2 AO 2 OM 2  Theorem of Pythagoras BO22  OM  AO  BO  r  BM 2  Theorem of Pythagoras AM BM Axiom 1: Assume the Theorem of Pythagoras. ABC with  B  90   AC2  AB 2  BC 2 Theorem 1: OM chord AB  AM  BM

Note: Examinable - A Proof using congruency can also be used Investigation : Line segment from centre of circle to midpoint of chord is perpendicular to the chord.

Many more investigations possible by means of GeoGebra. Suggested Conclusion : 

Conjecture 2 : AM MB  OM  chord AB Theorem : Line segment from centre of circle to midpoint of chord is perpendicular to the chord. Given : Any O with M the midpoint of chord AB Aim is to prove that : OM AB Construction : Draw AO , BO and OM

Proof: In  's OAM and OBM , AO BO  radii  Theorem 2 is the converse of Theorem 1. OM OM  common  OM  chord AB  AM  BM AM BM  given Axiom 2 : Assume two  's are congruent OAM   OBM s , s , s    2   1  90   AMB  180  Three sides of one triangle are equal OM AB to three sides of the other triangle Theorem 2 : AM MB  OM  chord AB Not Examinable as a Proof Investigation : Perpendicular bisector of a chord passes through the centre of a circle.

Suggested Conclusion :



Conjecture 3 : MN AB  AM  BM  O  MN Theorem : Perpendicular bisector of a chord passes through the centre of a circle. Given : Any O with MN a perpendicular bisector of chord AB Aim is to prove that : O MN Construction : Draw AN and NB Proof: In  's NAM and NBM , AMN   BMN 90   given  NMNM  common  NAM   NBM  Axiom: s , , s AM BM  given AN BN But OA OB r Any point on perpendicular bisector  OAB is equidistant from and   MN is equidistant from A and B . O MN  Theorem 3 : MN AB  AM  BM  O  MN

Not Examinable as a Proof Constructing the centre of any given circle

Theorem 3 can be utilized to construct the centre of any circle :

Given any Draw perpendicular bisectors to any two chords Centre is the point of intersection of these two perpendicular bisectors

Not Examinable Constructing the circumscribed circle about a given triangle

The point of concurrency of the perpendicular bisectors of the sides of a triangle is called the circumcentre of the triangle. The circle which passes through the vertices of the triangle is the circumcircle or circumscribed circle of the triangle. For enrichment Method : Draw perpendicular bisectors Why does this method work? DF and EG of AB and BC AOD   BOD  s ,  , s  AO  OB respectively. BOE   COE  s ,  , s  OB  OC DF EG O AO  OB  OC With O as centre and OB as Thus O and radius OB radius draw circumscribed circle. will pass through ABC , and . Find the value of x in each of the following riders

Solution : Rider 3 JL JK  r  x Solution : Rider 1 Solution : Rider 2 NJ JL  LN  x  2 AD bisects BC FG JN KM FH 4 BC 2 KJNJKN2  2  2  Pyth BD   4 FE EI  r 10 2 2 xx22  26  x2 AB 2 BD 2 (Pyth) x2 FE 2 FH 2  Pyth x22  x 4 x  4  36 22 22 x 5  4  3 x 10  4  2 21 4xx  40   10 Tutorial 1: Find the value of x in each of the following riders

PAUSE Unit

• Do Tutorial 1 • Then View Solutions Tutorial 1: Rider 1: Suggested Solution

Solution : Rider 1 BC BD 3 2

AB2 BD 2 AD 2  Pyth

x2 34 2  2 x 322  4  5 Tutorial 1: Rider 2: Suggested Solution

Solution : Rider 2 EJ2 FE 2  FJ 2   EFJ : Pyth EJ  FE22  FJ

22 FG  10  8  6 FJ   8 2 EK2 EI 2  KI 2   EKI : Pyth EK  EI22  KI EI r 10 22   10  5  5 3 HI KI 5 2 x  JK  EK  EJ 5 3  6 Tutorial 1: Rider 3: Suggested Solution

Solution : Rider 3 NP NR 12 2 NO NS 12.5 2 NM2 NR 2  MR 2   NRM : Pyth NM  r 1222  5  13 MS2 NM 2  NS 2   NSM : Pyth x  NM22  NS

2 51  132  12.5  2 More riders related to perpendicular bisectors of chords

For each of the two additional riders : Make an appropriate sketch Prove each by utilizing Theorems 1 to 3 and other axioms

Rider 4 : O is the centre of two concentric circles. Chord AB of the greater circle cuts the smaller circle at C and D . Prove that AC DB .

Rider 5 : Two circles with centres MNAB and intersect at and . PQ with A PQ is parallel to MN , with P and Q on the two circles. Prove that PQ 2 MN . Rider 4: Suggested Solution

Rider 4 : Concentric circles are O is the centre of two concentric circles. 's with common centre. Chord AB of the greater circle cuts the smaller circle at C and D . Prove that AC DB .

Construction : Draw OE CD

Proof: AE BE  CE bisects chord AB CE DE  OE bisects chord CD

AE  CE  BE  DE AC DB Rider 5: Suggested Solution

Rider 5 : Two circles with centres MNAB and intersect at and . PQ with A PQ is parallel to MN , with P and Q on the two circles. Prove that PQ 2 MN . Proof: AP Construction : Draw MR AP AR  MR bisects chord AP and NS AQ 2 AQ AS  NS bisects chord AQ 2 MN RS  Opposite sides of m RMNS 

But MN RS  AR  AS AP AQ AP AQ PQ MN  AR  AS     2 2 2 2 PQ2 MN Tutorial 2: Riders linked to perpendicular bisectors of chords

For each of the two riders : Make an appropriate sketch Prove each by utilizing Theorems 1 to 3 and other axioms Rider 1: Prove that equal chords are equidistant from the centre of a circle. Rider 2 : AB is a chord of M and AC with B AC is drawn such that BC 2 AB . PAUSE Unit 3BC 2 Prove that MC22 MB . • Do Tutorial 2 2 • Then View Solutions Tutorial 2: Rider 1: Suggested Solution Rider 1: Prove that equal chords are equidistant from the centre of a circle.

Given : O with AB DE Proof: Construction : Draw OC AB AB BC  OC bisects chord AB and and CF DE 2 Aim is to prove that : OC OF DE FE  OF bisects chord DE 2 AB DE BC    FE  AB  DE 22 r 2 2 2 r OC r  BC   OCB : Pyth r22  FE  BC  FE OF2  OFE : Pyth OC  OF  OC22  OF  Tutorial 2: Rider 2: Suggested Solution

Rider 2 : Construction : Draw MD AB , ABMACBAC is a chord of and with  AM , MB and MC . is drawn such that 2.BCAB 3BC 2 Prove that .MCMB22 2 Proof: MC2 MD 2  DC 2   MDC : Pyth 1 MD2 MB 2  DB 2   MDB : Pyth 2 Substituting 4 and 5 into 3: 22 2 2 2 2 MC  MB  DB  DC From 1 and 2 3 22BC  5 BC    MC MB      AB 44    DB  MD bisects chord AB BC2225 BC 2 MB2   BC1 BC BC 16 16    BC  2 AB  AB  4 24BC 2 2 2 4 2 MB2 and DC DB BC  Obvious 16 2 BC5 BC 2 3BC  BC From 4  5 MB 44  2 Unit 2

Inscribed and Grade 11 Central Angle CAPS Theorems Mathematics Series Outcomes for Unit 2 In this Unit session we will :

• Investigate and prove that : Angle subtended by an arc (or chord) at the centre of a circle is double the size of the angle subtended by the same arc at the circumfrence of the circle. (Theorem 4) Alternative formulation : The central angle is double the inscribed angle subtended by the same chord.

• Investigate and prove that : Angles subtended by a chord at the circumference of a circle, on the same side of the chord, are equal. (Theorem 5)

• Solve riders related to Theorems 4 and 5. Inscribed and central angle terminology

An angle such as C in the accompanying sketch, whose vertex lies on a circle and whose sides are chords of a circle, is called an inscribed angle in the circle.

We also say that the major arc AB or the chord AB subtends ACB on the circle.

AOB whose vertex is the centre of the circle, is called a central angle of the circle. (Note sides of this angle are radii)

We say that the major arc AB or the chord AB subtends AOB at the centre. Investigation : The central angle of a circle is double the inscribed angle subtended by the same chord. Investigation : The central angle of a circle is double the inscribed angle subtended by the same chord.

Suggested Conclusion:

Conjecture : O with inscribed  BAC and central BOC  BOC  2  BAC Theorem : The central angle of a circle is double the inscribed angle subtended by the same chord. Given : O with inscribed  BAC and central BOC ProveAim is thatto prove : thatO : withBOC inscribed 2  BAC  BAC and central BOC Construction : Draw ADBOC with O  2 AD BAC Proof: 1   2  AO  OB ; radii of  BOD  1   2  ext.   sum of int. opp. s   2 1 Similarly: COD   3   4 Theorem : O with inscribed  BAC  2 3 and central BOC s ext.   sum of int. opp.    BOC  2  BAC BOC   BOD   COD  see sketch Examinable as a Proof  2  1  2  3  2  1   3  2 BAC Corollaries (Deductions) from Theorem 4

Corollary 1: The angle in a semicircle is a right angle. BOC 180 A   90  22

Corllary 2 : Chords are equal if they subtend equal angles at points on a circle. Proof will be discussed as an exercise.

Corollary 3 : Equal chords subtend equal inscribed angles at points on the same circle. Proof left as an exer cise.

Corollary 4 : Equal chords in the same circle subtend equal central angles. Proof left as exercise. Simple Riders linked to Theorem 4

Determine the value of angles represented by Greek letters. Solution : Rider 2 180  83    48.5  EF  EG 2 83   41.5    @ E=2   @H 2

Solution : Rider 3  50  Reason? Solution : Rider 1  130  Reason? ABC   ACB 44   AB  AC  180   2  44   92   Sum of  angles   46    2 Proof of Corollary 2 : Chords are equal if they subtend equal inscribed angles

Given : O with  BAC   EDF Aim is to prove that : BC EF Construction : Join BCEFO , , and to

Proof: BOC 2  A and  EOF  2  D  central   2  inscribed  BOC   EOF  Given that  A   D In BOC and  EOF :  BOC   EOF  proved OB OE  radii Corollaries 3 and 4 may be proved in the same way. OC OF radii   Proofs left as an exercise! BOC   EOF  s ,  , s BC EF Not Examinable as a Proof Tutorial 3: Central   2  inscribed 

1. Determine the value of the angles represented by the GREEK letters.

2. XY and ST are two parallel chords of M . PAUSE Unit XT YS P Prove that XMS   XPS . • Do Tutorial 3 3. AB is a diameter of O and AC a chord. • Then View Solutions The circle on AO as diameter cuts AC at M . BC Prove that MO  . 2 Tutorial 3: Problem 1: Rider 1: Suggested Solution

1. Determine the value of r .

CAB 2  D  central   2  inscribed 

rr 46   2 2rr   46  2rr   46  r 46  Tutorial 3: Problem 1: Rider 2: Suggested Solution

1. Determine the value of the angles represented by Greek letters.

GEF 180  GEN   NEF  180  180  98 82

GEF   41   central GEF  2  inscribed angle 2

GEF   41   central GEF  2  inscribed angle 2 Tutorial 3: Problem 1: Rider 3: Suggested Solution

1. Determine the value of the angles represented by Greek letters.

Reflexive KIJ  360   obtuse  KIJ  360   110   250  Obtuse KIJ    central   2  inscribed angle 2 110   55  2 Reflexive KIJ    central   2  inscribed angle 2 250   125  2 Tutorial 3: Problem 2: Suggested Solution

2. XY and ST are two parallel chords of M. XT YS P Prove that XMS   XPS . YXP alt.  T  XY ST  3 XMS 2  T  central   2  inscribed angle 1 XMS 2  Y  central   2  inscribed angle 22 TYTY       2 XPS   Y   YXP  ext.  of  XYP  TT    from 2 and 3   2 T   XMS  from 1  Tutorial 3: Problem 3: Suggested Solution

3. AB is a diameter of O and AC a chord. The circle on AO as diameter cuts AC at M . BC Prove that MO  . 2 AMO   ACB 90   angles in semicircles OM sinA   OMA is a right-angled  AO CB and sinA   BCA is a right-angled  AB OM CB AO AO   OM  CB  CB   AO  BO AO AB AB 2AO BC OM 2 Investigation : Angles subtended by a chord (or arc) of a circle, on the same side of the chord, are equal.

GeoGebra for more!

Note : Angles on same chord on different sides of chord are supplementary. Suggested Conclusion: (More about this in next video) Conjecture : Inscribed angles on the same side of a chord are equal Theorem : Angles subtended by a chord (or arc) of a circle, on the same side of the chord, are equal. Given : O with inscribed  BAC and inscribed angle BDC on the same side of chord BC Aim is to prove that : BAC   BDC Construction : Draw OB and OC Proof: BOC BDC   central   2  inscribed  2 and BOC BAC   central   2  inscribed  2 BDC   BAC  transitive property for equality Theorem 5 : O with inscribed  BAC and inscribed BDC on the same side of chord BC  BAC   BDC Simple Riders linked to Theorem 5

1. Determine the value of the angles represented by Greek letters.

Solution : Rider 2  37  same chord JG  37   alternate s   37   same chord HF 

Solution : Rider 3:  34   's on PL Solution : Rider 1    48  angles on chord BC NPQ 180    104   sum  's in   35  angles on chord AD NPQ 180   34   104   42    NPM 42    's on NM Tutorial 4: Angles subtended by a chord at a point on circle are equal

1. Determine the value of the angles represented by letters.

PAUSE Unit

• Do Tutorial 4 • Then View Solutions Tutorial 4: Rider 1: Suggested Solution

Determine the value of angles represented by letters in the rider.

Solution : Rider 1 a40  angles on same chord AD b36  angles on same chord BC Tutorial 4: Rider 2: Suggested Solution

Determine the value of angles represented by letters in the rider.

Solution : Rider 2 c33  angles on same chord HF  ec 33   alternate angles d e 33   angles on same chord JG Tutorial 4: Rider 3: Suggested Solution

Determine the value of angles represented by letters in the rider.

Solution: Rider 3: f34  angles on same chord LP

g h 42  (angles on the same chord MN )

j49  angles on same chord ML h180   34   49   55  =42   sum of angles in  NMP k f g 34 42 76  ext.  of  sum of opp. int. angles End of the First Topic Slides on Circle Geometry REMEMBER! • Consult text-books and past exam papers and memos for additional examples. • Attempt as many as possible other similar examples on your own. • Compare your methods with those that were discussed in these Topic Slides. • Repeat this procedure until you are confident. • Do not forget: Practice makes perfect!