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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

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**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.

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**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.

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**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

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**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.

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**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.

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**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.

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**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.

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**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

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**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. )

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**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.

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**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.

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**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.

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**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

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**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)

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**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

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**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

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**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)

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**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) 2PBC (90 50) 180 PBC 20 (b) APC PBC PCB (ext. of D) 70

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**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.

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**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.

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**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

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**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.

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**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) (

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**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) (

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**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. (

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**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

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**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.

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**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

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**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

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**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.

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**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

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**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.

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**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

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**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.)

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**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

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**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)

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**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. )

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**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. )

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**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. )

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**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.

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**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.

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**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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**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.

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**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.

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**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

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**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

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**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.

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**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.

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**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

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**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

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**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.

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**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)

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**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 (

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**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

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**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 (

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**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

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**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

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**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.)

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**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.)

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