# Basic Properties of Circles (1)

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

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

7.1 Chords of a Circle A. Basic Terms 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.

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

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

7.1 Chords of a Circle B. Chords 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 DAOP and DBOP.

7.1 Chords of a Circle B. Chords a Circle
The converse of Theorem 7.1 is also true. 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) This theorem can be proved by considering DOAP and DOBP.

7.1 Chords of a Circle B. Chords 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.

Example 7.1T 7.1 Chords of a Circle Solution: B. Chords a Circle
In the figure, O is the centre of the circle. AP  PB  5 cm and OP  12 cm. Find PQ. Solution: ∵ AP  PB ∴ OP  AB (line from centre to mid-pt. of chord  chord) In DOAP, OA2  OP2  AP2 (Pyth. Theorem) OA  cm  13 cm OQ  OA (radii)  13 cm ∴ PQ  (13 – 12) cm  1 cm

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

Example 7.3T 7.1 Chords of a Circle Solution: B. Chords a Circle
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. Solution: (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. In DOCP, OC2  OP2  PC2 (Pyth. Theorem) r2  (r – 4)2  82 8r  80 r  10 ∴ The radius of the circle is 10 cm.

7.1 Chords of a Circle B. Chords a Circle
Properties about a perpendicular line from the centre to a chord: 2. Distance between Chords and Centre 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) This theorem can be proved by considering DOAP and DOCQ.

7.1 Chords of a Circle B. Chords a Circle
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) This theorem can be proved by considering DOAP and DOCQ.

Example 7.4T 7.1 Chords of a Circle Solution: B. Chords a Circle
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.) Solution: ∵ AB  CD ∴ OM  ON (equal chords, equidistant from centre) ∵ All of the interior angles of the quadrilateral OMPN are right angles and OM  ON. ∴ OMPN is a square. In DONP, OP2  ON2  NP2 (Pyth. Theorem)  2ON2 ON  cm  cm

7.2 Angles of a Circle A. The 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 q  2f. (Reference:  at the centre twice  at ⊙ce)

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

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

7.2 Angles of a Circle B. The Angle in a Semicircle
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)

Example 7.6T 7.2 Angles of a Circle Solution:
B. The Angle in a Semicircle Example 7.6T 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. Solution: (a) Since AP is a diameter, ACP  90. ( in semicircle) In DACB, ∵ AC  BC ∴ PAC  PBC (base s, isos. D) PAC  PBC  ACB  180 ( sum of D) 2PBC  (90  50)  180 PBC  20 (b) APC  PBC  PCB (ext.  of D)  70

7.2 Angles of a Circle C. Angle in the Same Segment 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.

7.2 Angles of a Circle C. Angle in the Same Segment
The angles in the same segment of a circle are equal. 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) This theorem can be proved by considering the angle at the centre.

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

7.3 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.

7.3 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) (

7.3 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, (Reference: equal arcs, equal chords) (

7.3 Relationship among the Chords, Arcs and Angles of a Circle
The above theorems are summarized in the following diagram: Equal Chords Equal Arcs Equal 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. (

7.3 Relationship among the Chords, Arcs and Angles of a Circle
Example 7.8T 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. ( Solution: (a) ∵ AB  BC  CD  DE  EF  FA ∴ AOB  BOC  COD  DOE  EOF  FOA (equal chords, equal s) ∴ AOB  360  6  60 (b) ∵ AB  BC  CD  DE  EF  FA ∴ AB  BC  CD  DE  EF  FA (equal chords, equal arcs) ( ∴ AB  (30  6) cm (  5 cm

7.3 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  q : f. (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.

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

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

7.3 Relationship among the Chords, Arcs and Angles of a Circle
5. Arcs Proportional to Angles 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) ( Notes: In a circle, chords are not proportional to the angles subtend at the circumference. This theorem can be proved by constructing the corresponding angles at the centre for each arcs.

7.3 Relationship among the Chords, Arcs and Angles of a Circle
Example 7.11T 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. ( Solution: (a) ∵ AOB  2  ACB ( at the centre twice  at ⊙ce) ∴ ACB  40 In DAOE, OEC  80  20 (ext.  of D)  100 In DBCE, OEC  ACB  CBD (ext.  of D) 100  40  CBD CBD  60 (b) AB : CD  ACB : CBD (arcs prop. to s at ⊙ce) ( 12 cm : CD  40 : 60 ∴ CD  18 cm

7.4 Basic Properties of a Cyclic Quadrilateral
A. Opposite Angles of a Cyclic Quadrilateral Cyclic quadrilateral: quadrilateral with all vertices lying on a circle 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.

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

7.4 Basic Properties of a Cyclic Quadrilateral
B. Exterior Angles 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: Theorem 7.14 The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle, that is, f  q. (Reference: ext. , cyclic quad.)

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

7.4 Basic Properties of a Cyclic Quadrilateral
C. Tests for Concyclic Points Points are said to be concyclic if they lie on the same circle. 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)

7.4 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.  )

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

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

Chapter Summary 7.1 Chords of a Circle
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.

Chapter Summary 7.2 Angles of a Circle
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.

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. (

7.4 Basic Properties of a Cyclic Quadrilateral
Chapter Summary 7.4 Basic Properties of a Cyclic Quadrilateral If ABCD is a cyclic quadrilateral, then (a) a  b  180 and (b) a  c.

Follow-up 7.1 7.1 Chords of a Circle Solution: B. Chords a Circle
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. Solution: ∵ AP  PB ∴ OP  AB (line from centre to mid-pt. of chord  chord) In DOAP, OA2  OP2  AP2 (Pyth. Theorem) OA  cm  13 cm ∴ The radius of the circle is 13 cm.

Follow-up 7.2 7.1 Chords of a Circle Solution: B. Chords a Circle
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. Solution: APO  AQO  90 (given) PAO  QAO (given) AO  AO (common side) ∴ DAOP  DAOQ (AAS) ∴ AP  AQ (corr. sides,  Ds) ∵ OP  AC (given) ∴ AP  PC (line from centre  chord bisects chord) Similarly, AQ  QB (line from centre  chord bisects chord) ∴ AC  AB

Follow-up 7.3 7.1 Chords of a Circle Solution: B. Chords 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.) Solution: (a) ∵ PB  PC ∴ AP  BC (line from centre to mid-pt. of chord  chord) In DABP, AB2  AP2  BP2 (Pyth. Theorem) AP  cm  12 cm

Follow-up 7.3 7.1 Chords of a Circle Solution: B. Chords 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.) Solution: (b) Let r cm be the radius of the circle. Then OB  r cm and OP  (12 – r) cm. In DOBP, OB2  OP2  BP2 (Pyth. Theorem) r2  (12 – r)2  52 24r  169 r  ∴ The radius of the circle is cm.

Follow-up 7.4 7.1 Chords of a Circle Solution: B. Chords 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. Solution: ∵ AB  CD ∴ OQ  OP (equal chords, equidistant from centre) Consider the quadrilateral OPRQ. ∵ AP  PB and CQ  QD ∴ OP  AB and OQ  CD (line from centre to mid-pt. of chord  chord) ∴ OPA  OQC  90 ∵ PRQ  90 (given) ∴ POQ  360 – 90 – 90 – 90  90 ∴ OPRQ is a square.

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

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

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

7.3 Relationship among the Chords, Arcs and Angles of a Circle
Follow-up 7.8 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. ( Solution: (a) ∵ CD  AB ( ∴ CD  AB (equal arcs, equal chords)  9 cm (b) ∵ CD  AB ( ∴ COD  AOB (equal arcs, equal s)  78 In DCOD, ∵ OC  OD (radii) ∴ By considering the angle sum of DCOD, we have OCD  51. ∴ OCD  ODC (base s, isos. D)

7.3 Relationship among the Chords, Arcs and Angles of a Circle
Follow-up 7.9 In the figure, AOB  120, BOC  150 and ABC  54 cm. Find the length BC. ( Solution: Reflex AOC  270 BOC : Reflex AOC  BC : ABC (arcs prop. to s at centre) ( 150 : 270  BC : 54 cm ( BC  30 cm (

7.3 Relationship among the Chords, Arcs and Angles of a Circle
Follow-up 7.10 In the figure, O is the centre of the circle. AOB  100, AB  AD  CD. Find ABC : CD. ( Solution: ∵ CD  DA  AB ∴ COD  DOA  AOB (equal chords, equal s)  100 ∴ AOC  360  100  100 (s at a pt.)  160 ∴ ABC : CD  160 : 100 (arcs prop. to s at centre) (  8 : 5

7.3 Relationship among the Chords, Arcs and Angles of a Circle
Follow-up 7.11 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. ( Solution: (a) ∵ BOC  2  BAC ( at the centre twice  at ⊙ce) ∴ BAC  36 (b) BC : AD  BAC : ABD (arcs prop. to s at ⊙ce) ( BC : 12 cm  36 : 48 ( ∴ BC  9 cm (

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

Follow-up 7.13 7.4 Basic Properties of a Cyclic Quadrilateral
B. Exterior Angles of a Cyclic Quadrilateral Follow-up 7.13 In the figure, ABP, DCP and BCQ are straight lines. DAB  50 and DQC  45. (a) Find QDC. (b) Find APD. Solution: (a) In DABQ, ABQ  50  45  180 ( sum of D) ABQ  85 ∴ QDC  ABQ (ext. , cyclic quad.)  85 (b) In DADP, QDC  APD  50 (ext.  of D) 85  APD  50 APD  35

Follow-up 7.14 7.4 Basic Properties of a Cyclic Quadrilateral
C. Tests for Concyclic Points Follow-up 7.14 In the figure, AB // PQ, show that A, B, C and D are concyclic. Solution: PQC  ABQ (corr. s, AB // PQ) PQC  CDP  180 (opp. s, cyclic quad.) ∴ ABQ  CDP  180 i.e., ABC  CDA  180 ∴ A, B, C and D are concyclic. (opp. s supp.)

Follow-up 7.15 7.4 Basic Properties of a Cyclic Quadrilateral
C. Tests for Concyclic Points Follow-up 7.15 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. Solution: (a) In DADP, DAP  45  85  180 ( sum of D) DAP  50 (b) ∵ DAB  BCD  (50  50)  80  180 ∴ A, B, C and D are concyclic. (opp. s supp.)