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7 7.1Chords of a Circle 7.2Angles of a Circle Chapter Summary Case Study Basic Properties of Circles (1) 7.3Relationship among the Chords, Arcs and Angles of a Circle 7.4Basic Properties of a Cyclic Quadrilateral

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P. 2 In order to find the centre of the circular plate: Step 1: Draw an arbitrary triangle inscribed in the circular plate. Case Study You need to find the centre of the circular plate first. I found a fragment of a circular plate. How can I know its original size? Step 2:Find the circumcentre of the triangle, i.e., the centre of the circular plate, by drawing 3 perpendicular bisectors.

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P. 3 A. Basic Terms of a Circle 7.1 Chords of a Circle Circle:closed curve in a plane where every point on the curve is equidistant from a fixed point. Centre:fixed point Circumference:curve or the length of the curve Chord:line segment with two end points on the circumference Radius:line segment joining the centre to any point on the circumference Diameter:chord passing through the centre Remarks: 1.The length of a radius is half that of a diameter. 2.A diameter is the longest chord in a circle.

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P. 4 A. Basic Terms of a Circle 7.1 Chords of a Circle Arc:portion of the circumference Angle at the centre:angle subtended by an arc or a chord at the centre minor arc (e.g. AYB) shorter than half of the circumference ( major arc (e.g. AXB) longer than half of the circumference (

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P. 5 B. Chords a Circle 7.1 Chords of a Circle If we draw a chord AB on a circle and fold the paper as shown below: Then the crease passes through the centre of the circle; bisects the chord AB. is perpendicular to the chord AB; A, B coincide

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P. 6 B. Chords a Circle 7.1 Chords of a Circle Properties about a perpendicular line from the centre to a chord: 1.Perpendicular Line from Centre to a Chord Theorem 7.1 If a perpendicular line is drawn from a centre of a circle to a chord, then it bisects the chord. In other words,if OP AB, then AP BP. (Reference: line from centre chord bisects chord) This theorem can be proved by considering AOP and BOP.

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P. 7 B. Chords a Circle 7.1 Chords of a Circle This theorem can be proved by considering OAP and OBP. Theorem 7.2 If a line is joined from the centre of a circle to the mid-point of a chord, then it is perpendicular to the chord. In other words,if AP BP, then OP AB. (Reference: line from centre to mid-pt. of chord chord) The converse of Theorem 7.1 is also true.

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P. 8 B. Chords a Circle 7.1 Chords of a Circle From Theorem 7.1 and Theorem 7.2, we obtain an important property of chords: The perpendicular bisector of any chord of a circle passes through the centre.

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P. 9 Example 7.1T Solution: In the figure, O is the centre of the circle. AP PB 5 cm and OP 12 cm. Find PQ. B. Chords a Circle 7.1 Chords of a Circle ∵ AP PB ∴ OP AB (line from centre to mid-pt. of chord chord) In OAP, OA 2 OP 2 AP 2 (Pyth. Theorem) OA cm 13 cm OQ OA (radii) 13 cm ∴ PQ (13 – 12) cm 1 cm

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P. 10 Example 7.2T Solution: In the figure, O is the centre of the circle. AOB is a straight line and OM BC. Show that ABC OBM. B. Chords a Circle 7.1 Chords of a Circle ∵ OM BC ∴ BM MC (line from centre chord bisects chord) ∴ BC : BM 2 : 1 ∵ OB OA (radii) ∴ AB : OB 2 : 1 OBM ABC (common ) ∴ ABC OBM (ratio of 2 sides, inc. )

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P. 11 Example 7.3T Solution: In the figure, O is the centre and AB is a diameter of the circle. AB CD, PB 4 cm and CD 16 cm. (a)Find the length of PC. (b)Find the radius of the circle. B. Chords a Circle 7.1 Chords of a Circle In OCP, OC 2 OP 2 PC 2 (Pyth. Theorem) (a) ∵ OB CD ∴ PC PD (line from centre chord bisects chord) 8 cm (b)Let r cm be the radius of the circle. Then OC r cm and OP (r – 4) cm. r 2 (r – 4) 2 8 2 8r 80 r 10 ∴ The radius of the circle is 10 cm.

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P. 12 B. Chords a Circle 7.1 Chords of a Circle Properties about a perpendicular line from the centre to a chord: 2.Distance between Chords and Centre This theorem can be proved by considering OAP and OCQ. Theorem 7.3 If the lengths of two chords are equal, then they are equidistant from the centre. In other words,if AB CD, then OP OQ. (Reference: equal chords, equidistant from centre)

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P. 13 B. Chords a Circle 7.1 Chords of a Circle This theorem can be proved by considering OAP and OCQ. The converse of Theorem 7.3 is also true. Theorem 7.4 If two chords are equidistant from the centre of a circle, then their lengths are equal. In other words,if OP OQ, then AB CD. (Reference: chords equidistant from centre are equal)

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P. 14 Example 7.4T Solution: In the figure, O is the centre of the circle. AB CD, AB CD, OM AB and ON CD. If OP 6 cm, find ON. (Give the answer in surd form.) B. Chords a Circle 7.1 Chords of a Circle ∵ AB CD ∴ OM ON (equal chords, equidistant from centre) ON cm ∵ All of the interior angles of the quadrilateral OMPN are right angles and OM ON. ∴ OMPN is a square. In ONP, OP 2 ON 2 NP 2 (Pyth. Theorem) 2ON 2 cm

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P. 15 A. The Angle at the Circumference 7.2 Angles of a Circle Angle at the circumference: angle subtended by an arc (or a chord) at the circumference Angle at the centre: angle subtended by an arc (or a chord) at the centre Relationship between these angles: Theorem 7.5 In each of the above figures, the angle at the centre subtended by an arc is twice the angle at the circumference subtended by the same arc. This means that 2 . (Reference: at the centre twice at ⊙ ce )

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P. 16 A. The Angle at the Circumference 7.2 Angles of a Circle This theorem can be proved by constructing a diameter PQ. Since OA OP (radii), AOP is isosceles. ∴ OAP OPA a. Hence the exterior angle of AOQ 2a. In the left semicircle: Similarly, in the right semicircle, BOQ 2b. ∵ 2a 2b and a b ∴ 2

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P. 17 Example 7.5T Solution: In the figure, AB and CD are two parallel chords of the circle with centre O. BOD 70 and MDO 10 . Find ODC. ∵ BOD 2 BAD ( at the centre twice at ⊙ ce ) ∴ BAD 35 A. The Angle at the Circumference 7.2 Angles of a Circle ODC 10 BAD (alt. s, AB // CD) ODC 10 35 ODC 25

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P. 18 B. The Angle in a Semicircle 7.2 Angles of a Circle In the figure, if AB is a diameter of the circle with centre O, then the arc APB is a semicircle and APB is called the angle in a semicircle. Since the angle at the centre AOB 180 , the angle at the circumference APB 90 . ( at the centre twice at ⊙ ce ) Theorem 7.6 The angle in a semicircle is 90 . That is,if AB is a diameter, then APB 90 . (Reference: in semicircle) Conversely,if APB 90 , then AB is a diameter. (Reference: converse of in semicircle)

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P. 19 Example 7.6T Solution: In the figure, AP is a diameter of the circle with centre O and AC BC. If PCB 50 , find (a) PBC and (b) APC. (a)Since AP is a diameter, ACP 90 . ( in semicircle) 7.2 Angles of a Circle PBC 20 B. The Angle in a Semicircle ∵ AC BC ∴ PAC PBC (base s, isos. ) In ACB, PAC PBC ACB 180 ( sum of ) 2 PBC (90 50 ) 180 (b) APC PBC PCB (ext. of ) 70

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P. 20 C. Angle in the Same Segment 7.2 Angles of a Circle Segment: region enclosed by a chord and the corresponding arc subtended by the chord Major segment APB area greater than half of the circle Minor segment AQB area less than half of the circle Angles in the same segment: angles subtended on the same side of a chord at the circumference Notes: We can construct infinity many angles in the same segment.

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P. 21 C. Angle in the Same Segment 7.2 Angles of a Circle Theorem 7.7 The angles in the same segment of a circle are equal, that is,if AB is a chord, then APB AQB. (Reference: s in the same segment) The angles in the same segment of a circle are equal. This theorem can be proved by considering the angle at the centre.

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P. 22 Example 7.7T Solution: In the figure, AC and BD are two chords that intersect at P. (a)Show that ABP DCP. (b)If AP 8, BP 12 and PC 6, find PD. (a)In ABP and DCP, 7.2 Angles of a Circle A D ( s in the same segment) (b) ∵ ABP DCP PD 4 C. Angle in the Same Segment B C ( s in the same segment) APB DPC (vert. opp. s) ∴ ABP DCP (AAA) ∴ (corr. sides, s)

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P Relationship among the Chords, Arcs and Angles of a Circle 1.Equal Chords and Equal Angles at the Centre Theorem 7.8 In a circle, if the angles at the centre are equal, then they stand on equal chords, that is, if x y, then AB CD. (Reference: equal s, equal chords) Conversely, equal chords in a circle subtend equal angles at the centre, that is, if AB CD, then x y. (Reference: equal chords, equal s) This theorem can be proved using congruent triangles.

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P Relationship among the Chords, Arcs and Angles of a Circle 2.Equal Angles at the Centre and Equal Arcs Theorem 7.9 In a circle, if the angles at the centre are equal, then they stand on equal arcs, that is, if p q, then AB CD. (Reference: equal s, equal arcs) Conversely, equal arcs in a circle subtend equal angles at the centre, that is, if AB CD, then p q. (Reference: equal arcs, equal s) (( ((

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P Relationship among the Chords, Arcs and Angles of a Circle 3.Equal Chords and Equal Arcs Theorem 7.10 In a circle, equal chords cut arcs with equal lengths, that is, if AB CD, then AB CD. (Reference: equal chords, equal arcs) Conversely, equal arcs in a circle subtend equal chords, that is, if AB CD, then AB CD. (Reference: equal arcs, equal chords) (( ((

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P Relationship among the Chords, Arcs and Angles of a Circle The above theorems are summarized in the following diagram: Equal Chords Equal ArcsEqual Angles Theorem 7.10 Theorem 7.8 Theorem 7.9 Example: In the figure, the chords AB, BC and CA are of the same length. ∴ Each of the angles at the centre are equal, i.e., AOB BOC COA 120 . ∴ Each of the arcs are equal, i.e., AB BC CA 9 cm. (((

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P. 27 Example 7.8T Solution: (a) ∵ AB BC CD DE EF FA ∴ AOB BOC COD DOE EOF FOA (equal chords, equal s) (b) ∵ AB BC CD DE EF FA 7.3 Relationship among the Chords, Arcs and Angles of a Circle In the figure, O is the centre of the circle with circumference 30 cm. A regular hexagon ABCDEF is inscribed in the circle. (a)Find AOB. (b)Find the length of AB. ( ∴ AOB 360 6 ∴ AB BC CD DE EF FA (equal chords, equal arcs) (((((( ∴ AB (30 6) cm ( 5 cm 60

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P Relationship among the Chords, Arcs and Angles of a Circle 4.Arcs Proportional to Angles at the Centre Theorem 7.11 In a circle, arcs are proportional to the angles at the centre, that is, AB : PQ : . (Reference: arcs prop. to s at centre) (( Notes: 1.In a circle, chords are not proportional to the angles subtend at the centre. 2.In a circle, chords are not proportional to the arcs.

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P. 29 Example 7.9T Solution: 7.3 Relationship among the Chords, Arcs and Angles of a Circle In the figure, O is the centre of the circle. APB 15 cm, PB 6 cm and POB 80 . Find AOP. ( ( AP 9 cm ( AOP : POB AP : PB (arcs prop. to s at centre) (( AOP : 80 9 cm : 6 cm AOP 120

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P. 30 Example 7.10T Solution: 7.3 Relationship among the Chords, Arcs and Angles of a Circle AOB : BOC AB : BC (arcs prop. to s at centre) (( 40 : BOC 2 : 3 BOC 60 In the figure, O is the centre of the circle. 3AB 2BC and AOB 40 . Find ABC : AEDC. ( ( ( ∵ 3AB 2BC (( ∴ AB : BC 2 : 3 (( ∴ AOC 100 and Reflex AOC 260 5 : 13 ∴ ABC : AEDC 100 : 260 (arcs prop. to s at centre) (

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P Relationship among the Chords, Arcs and Angles of a Circle 5.Arcs Proportional to Angles at the Circumference Notes: In a circle, chords are not proportional to the angles subtend at the circumference. Theorem 7.12 In a circle, arcs are proportional to the angles subtended at the circumference, that is, AB : PQ a : b. (Reference: arcs prop. to s at ⊙ ce ) (( This theorem can be proved by constructing the corresponding angles at the centre for each arcs.

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P. 32 Example 7.11T Solution: 7.3 Relationship among the Chords, Arcs and Angles of a Circle In AOE, OEC 80 20 (ext. of ) In the figure, O is the centre of the circle. AOB 80 , OAC 20 and AB 12 cm. (a)Find CBD. (b)Find the length of CD. ( ( (a) ∵ AOB 2 ACB ( at the centre twice at ⊙ ce ) ∴ ACB 40 100 In BCE, OEC ACB CBD (ext. of ) 100 40 CBD CBD 60 12 cm : CD 40 : 60 (b) AB : CD ACB : CBD (arcs prop. to s at ⊙ ce ) (( ∴ CD 18 cm

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P Basic Properties of a Cyclic Quadrilateral Cyclic quadrilateral: quadrilateral with all vertices lying on a circle A. Opposite Angles of a Cyclic Quadrilateral Two pairs of opposite angles: BAD and DCB ABC and CDA Theorem 7.13 The opposite angles in a cyclic quadrilateral are supplementary. Symbolically, BAD DCB 180 and ABC CDA 180 . (Reference: opp. s, cyclic quad.) This theorem can be proved by constructing the corresponding angles at the centre.

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P. 34 Example 7.12T Solution: In the figure, ABCD is a cyclic quadrilateral. AD is a diameter of the circle and DAC 35 . Find ABC. 7.4 Basic Properties of a Cyclic Quadrilateral A. Opposite Angles of a Cyclic Quadrilateral ∵ AD is a diameter. ∴ ACD 90 ( in semicircle) In ACD, 35 ACD ADC 180 ( sum of ) ∴ ABC ADC 180 (opp. s, cyclic quad.) 35 90 ADC 180 ADC 55 ABC 125

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P Basic Properties of a Cyclic Quadrilateral From Theorem 7.13, we obtain the following relationship between the exterior angle and the interior opposite angle of a cyclic quadrilateral: B. Exterior Angles of a Cyclic Quadrilateral Theorem 7.14 The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle, that is, . (Reference: ext. , cyclic quad.)

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P. 36 Example 7.13T Solution: In the figure, two circles meet at C and D. ADE and BCF are straight lines. If BAD 105 , find DEF. 7.4 Basic Properties of a Cyclic Quadrilateral FCD BAD (ext. , cyclic quad.) B. Exterior Angles of a Cyclic Quadrilateral 105 ∴ DEF FCD 180 (opp. s, cyclic quad.) DEF 75

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P Basic Properties of a Cyclic Quadrilateral Points are said to be concyclic if they lie on the same circle. C. Tests for Concyclic Points To test whether a given set of 4 points are concyclic (or a given quadrilateral is cyclic): Theorem 7.15 (Converse of Theorem 7.7) In the figure,if p q, then A, B, C and D are concyclic. (Reference: converse of s in the same segment)

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P Basic Properties of a Cyclic Quadrilateral C. Tests for Concyclic Points Theorem 7.16 (Converse of Theorem 7.13) In the figure,if a c 180 (or b d 180 ), then A, B, C and D are concyclic. (Reference: opp. s supp.) Theorem 7.17 (Converse of Theorem 7.14) In the figure,if p q, then A, B, C and D are concyclic. (Reference: ext. int. opp. )

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P. 39 Example 7.14T Solution: In the figure, APB and RDQC are straight lines. If AD // PQ, show that P, Q, C and B are concyclic. 7.4 Basic Properties of a Cyclic Quadrilateral ADR ABC (ext. , cyclic quad.) ∴ ABC PQR C. Tests for Concyclic Points ADR PQR (corr. s, AD // PQ) ∴ P, Q, C and B are concyclic. (ext. int. opp. )

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P. 40 Example 7.15T Solution: Consider the cyclic quadrilateral PQCD. (a)Find y. (b)Write down another four concyclic points. 7.4 Basic Properties of a Cyclic Quadrilateral C. Tests for Concyclic Points (a)y 110 180 (opp. s, cyclic quad.) (b) ∵ ABQ QPD 70 ∴ A, B, Q and P are concyclic. (ext. int. opp. ) y 70

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P Chords of a Circle Chapter Summary 1.If a perpendicular line is drawn from the centre of the circle to a chord, then it bisects the chord, and vice versa. 2.If the lengths of two chords are equal, then they are equidistant from the centre of the circle, and vice versa.

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P Angles of a Circle Chapter Summary 1.The angle at the centre is twice the angle at the circumference subtended by the same arc, that is, x 2y. 2.If AB is a diameter, then APB 90 . Conversely, if the angle at the circumference APB 90 , then AB is a diameter. 3.The angles in the same segment are equal, that is, x y.

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P. 43 Chapter Summary 7.3 Relationship among the Chords, Arcs and Angles of a Circle 1.Equal angles at the centre stand on equal chords. 2.Equal angles at the centre stand on equal arcs. 3.Equal arcs subtend equal chords. 4.Arcs are proportional to the angles at the centre. AB : PQ x : y (( 5.The arcs are proportional to the angles subtended at the circumference, that is, AB : BC x : y. ((

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P Basic Properties of a Cyclic Quadrilateral Chapter Summary If ABCD is a cyclic quadrilateral, then (a)a b 180 and (b)a c.

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Follow-up 7.1 Solution: In the figure, O is the centre of the circle. AB 24 cm and OP 5 cm. P is the mid-point of AB. Find the radius of the circle. ∵ AP PB ∴ OP AB (line from centre to mid-pt. of chord chord) In OAP, OA 2 OP 2 AP 2 (Pyth. Theorem) OA cm 13 cm ∴ The radius of the circle is 13 cm. B. Chords a Circle 7.1 Chords of a Circle

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Follow-up 7.2 Solution: In the figure, O is the centre of the circle. APC and AQB are straight lines. OP AC, OQ AB and OA is the angle bisector of CAB. Show that AC AB. APO AQO 90 (given) PAO QAO (given) ∴ AP AQ (corr. sides, s) ∵ OP AC (given) B. Chords a Circle 7.1 Chords of a Circle AO AO (common side) ∴ AOP AOQ (AAS) ∴ AP PC (line from centre chord bisects chord) Similarly, AQ QB (line from centre chord bisects chord) ∴ AC AB

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Follow-up 7.3 Solution: B. Chords a Circle 7.1 Chords of a Circle In the figure, O is the centre of the circle. BP CP 5 cm and AB 13 cm. (a)Find the length of AP. (b)Find the radius of the circle. (Give the answer in fraction.) (a) ∵ PB PC ∴ AP BC (line from centre to mid-pt. of chord chord) In ABP, AB 2 AP 2 BP 2 (Pyth. Theorem) AP cm 12 cm

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Follow-up 7.3 Solution: B. Chords a Circle 7.1 Chords of a Circle In the figure, O is the centre of the circle. BP CP 5 cm and AB 13 cm. (a)Find the length of AP. (b)Find the radius of the circle. (Give the answer in fraction.) (b)Let r cm be the radius of the circle. In OBP, OB 2 OP 2 BP 2 (Pyth. Theorem) Then OB r cm and OP (12 – r) cm. r 2 (12 – r) 2 r 169 r ∴ The radius of the circle is cm.

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Follow-up 7.4 Solution: B. Chords a Circle 7.1 Chords of a Circle In the figure, O is the centre of the circle and P and Q are the mid-points of AB and CD respectively. If AB CD and AB CD, show that OPRQ is a square. ∵ AB CD ∴ OQ OP (equal chords, equidistant from centre) Consider the quadrilateral OPRQ. ∴ OPRQ is a square. ∵ AP PBand CQ QD ∴ OP ABand OQ CD (line from centre to mid-pt. of chord chord) ∴ OPA OQC 90 ∵ PRQ 90 (given) ∴ POQ 360 – 90 – 90 – 90 90

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Follow-up 7.5 Solution: A. The Angle at the Circumference 7.2 Angles of a Circle In the figure, O is the centre of the circle. AB // DC, CDP = 30 and ACB = 90 . Find ABC. AOD ODC (alt. s, AB // DC) 30 ABC 75 ∵ AOD 2 ACD ( at the centre twice at ⊙ ce ) ∴ ACD 15 BAC ACD (alt. s, AB // DC) 15 ∴ In ABC, ABC BAC 90 180 ( sum of ) ABC 15 90 180

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Follow-up 7.6 Solution: B. The Angle in a Semicircle 7.2 Angles of a Circle The figure shows a circle with diameter AB and AFB 120 . Find EDC. Since AP is a diameter, ACB 90 and AEB 90 . ( in semicircle) EDC 60 In the quadrilateral DEFC, DEF 90 (adj. s on st. line) DCF 90 (adj. s on st. line) EFC 120 (vert opp. s) EDC DEF EFC DCF 360 ( sum of polygon) EDC 90 120 90 360

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Follow-up 7.7 Solution: 7.2 Angles of a Circle In the figure, PB AB and ABP 36 . (a)Find BAP and BDC. (b)Show that CDP is an isosceles triangle. C. Angle in the Same Segment (a)In ABP, ∵ AB PB (given) (b)From (a), BPA BAP 72 and BDC 72 . ∴ BAP BPA (base s, isos s) BAP BPA 36 180 ( sum of ) BAP 72 BDC BAP 72 ( s in the same segment) ∵ DPC BPA 72 (vert opp. s) ∴ DPC PDC 72 ∴ CP CD (sides opp. equal s) i.e., CDP is an isosceles triangle.

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Follow-up Relationship among the Chords, Arcs and Angles of a Circle Solution: ∴ CD AB (equal arcs, equal chords) In the figure, O is the centre of the circle. AB CD, AB 9 cm and AOB 78 . (a)Find the length of CD. (b)Find OCD. (( 9 cm (a) ∵ CD AB (( (b) ∵ CD AB (( ∴ COD AOB (equal arcs, equal s) 78 In COD, ∵ OC OD (radii) ∴ OCD ODC (base s, isos. ) ∴ By considering the angle sum of COD, we have OCD 51 .

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Follow-up Relationship among the Chords, Arcs and Angles of a Circle Solution: In the figure, AOB 120 , BOC 150 and ABC 54 cm. Find the length BC. ( ( Reflex AOC 270 BOC : Reflex AOC BC : ABC (arcs prop. to s at centre) ( ( 150 : 270 BC : 54 cm ( BC 30 cm (

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Follow-up Relationship among the Chords, Arcs and Angles of a Circle Solution: ∵ CD DA AB ∴ COD DOA AOB (equal chords, equal s) ∴ AOC 360 100 100 ( s at a pt.) 8 : 5 In the figure, O is the centre of the circle. AOB 100 , AB AD CD. Find ABC : CD. ( ( 100 160 ∴ ABC : CD 160 : 100 (arcs prop. to s at centre) ( (

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Follow-up Relationship among the Chords, Arcs and Angles of a Circle In the figure, O is the centre of the circle. ABD 48 , BOC 72 and AD 12 cm. (a)Find BAC. (b)Find the length of BC. ( ( (a) ∵ BOC 2 BAC ( at the centre twice at ⊙ ce ) (b) BC : AD BAC : ABD (arcs prop. to s at ⊙ ce ) (( Solution: BC : 12 cm 36 : 48 ( ∴ BC 9 cm ( ∴ BAC 36

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Follow-up 7.12 Solution: ∵ CD AB ∴ CBD ADB (equal chords, equal s) In the figure, ABCD is a cyclic quadrilateral. AB CD, ADB 46 and ABD 32 . Find CDB. 46 7.4 Basic Properties of a Cyclic Quadrilateral A. Opposite Angles of a Cyclic Quadrilateral ABC CDA 180 (opp. s, cyclic quad.) CDB 56 (32 46 ) ( CDB 46 ) 180

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Follow-up Basic Properties of a Cyclic Quadrilateral Solution: (a)In ABQ, ABQ 50 45 180 ( sum of ) In the figure, ABP, DCP and BCQ are straight lines. DAB 50 and DQC 45 . (a)Find QDC.(b)Find APD. B. Exterior Angles of a Cyclic Quadrilateral ABQ 85 ∴ QDC ABQ (ext. , cyclic quad.) 85 (b)In ADP, QDC APD 50 (ext. of ) 85 APD 50 APD 35

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Follow-up Basic Properties of a Cyclic Quadrilateral C. Tests for Concyclic Points Solution: In the figure, AB // PQ, show that A, B, C and D are concyclic. PQC ABQ (corr. s, AB // PQ) PQC CDP 180 (opp. s, cyclic quad.) ∴ A, B, C and D are concyclic. (opp. s supp.) ∴ ABQ CDP 180 i.e., ABC CDA 180

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Follow-up 7.15 Solution: 7.4 Basic Properties of a Cyclic Quadrilateral C. Tests for Concyclic Points In the figure, BAP 50 , APD 85 , ADP 45 and BCD 80 . (a)Find DAP. (b)Show that A, B, C and D are concyclic. (a)In ADP, DAP 45 85 180 ( sum of ) DAP 50 (b) ∵ DAB BCD (50 50 ) 80 180 ∴ A, B, C and D are concyclic. (opp. s supp.)

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