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: OMchord 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 NM NM 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 KJ2 NJ 2 KN 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 , AB is a chord of M and AC with B AC AM, MB and MC.