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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.

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Presentation on theme: "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."— Presentation transcript:

1 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

2 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.

3 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.

4 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 (

5 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

6 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.

7 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.

8 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.

9 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

10 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.  )

11 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.

12 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)

13 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)

14 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

15 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 )

16 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 

17 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 

18 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)

19 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 

20 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.

21 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.

22 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)

23 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.

24 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) (( ((

25 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) (( ((

26 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. (((

27 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 

28 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.

29 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 

30 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) ( 

31 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.

32 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

33 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.

34 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 

35 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.)

36 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 

37 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)

38 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.  )

39 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.  )

40 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 

41 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.

42 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.

43 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. ((

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

45 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

46 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

47 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

48 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.

49 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 

50 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 

51 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 

52 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.

53 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 .

54 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 (

55 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) ( (

56 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 

57 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 

58 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 

59 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 

60 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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