With the Educators, for the Educators

MATHEMATICS WORKSHOP

EUCLIDEAN GEOMETRY

TEXTBOOK GRADE 11 (Chapter 8)

Presented by: Jurg Basson

MIND ACTION SERIES

Attending this Workshop = 10 SACE Points

1

tangent

secant

diameter

chord

arc

radiussector

.

.segment

CHAPTER 8 EUCLIDEAN GEOMETRY BASIC CIRCLE TERMINOLOGY

THEOREMS INVOLVING THE CENTRE OF A CIRCLE THEOREM 1 A The line drawn from the centre of a circle perpendicular to a chord bisects the chord. (line from centre ⊥ to chord) If OM AB⊥ then AM MB= Proof Join OA and OB. In OAM and OBM:Δ Δ (a) OA OB= radii (b) 1 2

ˆ ˆM M 90= = ° given (c) OM OM= common

OAM OBM∴Δ ≡ Δ RHS AM MB∴ =

THEOREM 1A (Converse) The line segment joining the centre of a circle to the midpoint of a chord is perpendicular to the chord. (line from centre to midpt of chord) If AM MB= then OM AB⊥

Radius: A line from the centre to any point on the circumference of the circle. Chord: A line with end-points on the circumference. Diameter: A chord passing through the centre of the circle. It is double the length of the radius. Tangent: A line touching the circle at only one point. Secant: A line passing through two points on the circle.

= =

2

Definition The perpendicular bisector of a line is a line that bisects the given line at right angles. In the diagram, OM is the perpendicular bisector of AB. THEOREM 1B The perpendicular bisector of a chord passes through the centre of the circle. (perp bisector of chord) EXAMPLE 1 O is the centre. AB 8 cm, OF 3 cm, OE 4 cm= = = , AF FB= and CD OE⊥ . Calculate the length of chord CD. Solution

AF 4 cm= AB 8 cm= and AF FB= 1F 90∴ = ° line from centre to midpt of chord 2 2 2OA (3) (4)= + Pythagoras

OA 5 cm∴ = OD 5 cm∴ = equal radii

2 2 2ED (5) (4)= − Pythagoras; 1E 90= ° ED 3 cm∴ =

But DE CE= line from centre ⊥ to chord CD 6 cm∴ =

EXERCISE 1 In all questions, O is the centre. (a) Calculate the length of AC. (b) Calculate the length of DE. (c) Calculate the length of (d) Determine the radius OB in terms the radius of the circle. of x and hence the length of OB. and hence PQ.

3

=

=

8

A

EB

D x12

O.

3

(e) The radius of the semi-circle centre O is 5 cm. A square is fitted into the semi-circle as shown in the diagram. Calculate the area of the square. Let the length of the square equal x.

Subtended Angles

Arc APB or chord AB subtends ˆACB at the circumference of the circle and ˆAOB at the centre. THEOREM 2 The angle subtended by an arc at the centre of a circle is double the size of the angle subtended by the same arc at the circumference (on the same side of the chord as the centre). ( at centre 2 at circ∠ = ×∠ ) Proof Consider Diagram 1 and Diagram 2. Join CO and produce.

1 1ˆ ˆ ˆO C A= + ext of AOCΔ

Now OA OC= radii equal 1

ˆ ˆC A∴ = s∠ opp = sides

1 1ˆ ˆO 2C∴ =

Similarly, in OCBΔ , 2 2ˆ ˆO 2C=

1 2 1 2ˆ ˆ ˆ ˆO O 2C 2C∴ + = +

1 2 1 2ˆ ˆ ˆ ˆO O 2(C C )∴ + = +

ˆ ˆAOB 2ACB∴ = Consider Diagram 3. Join CO and produce.

1 1ˆ ˆ ˆO C A= + ext of AOCΔ

Now OA OC= radii equal 1

ˆ ˆC A∴ = s∠ opp = sides

1 1ˆ ˆO 2C∴ =

∠

∠

(f) AB is a chord of the circle. AC CB,= and the length of OA is 5 units. AC 3 units and OC 4 units= = .

Show that OC AB⊥ and explain why OC passes through the centre O.

.x

2x. x

2x

.x

2x

4

Similarly, in OCBΔ , 2 2ˆ ˆO 2C=

2 1 2 1ˆ ˆ ˆ ˆO O 2C 2C∴ − = −

2 1 2 1ˆ ˆ ˆ ˆO O 2(C C )∴ − = −

ˆ ˆAOB 2ACB∴ = THEOREM 3 The angle subtended at the circle by a diameter is a right angle. We say that the angle in a semi-circle is . ( ∠ in a semi-circle) If AB is a then ˆACB 90= ° diameter THEOREM 3 (Converse) If the angle subtended by a chord at a point on the circle is , then the chord is a diameter. (Chord subtends 90° ) If ˆACB 90= ° then AB is a diameter EXAMPLE 2 O is the centre of each circle. Calculate the sizes of the angles marked with small letters. Solutions

65a = ° at centre 2 at circ∠ = ×∠ 25b = ° at centre 2 at circ∠ = ×∠

1O 260= ° at centre 2 at circ∠ = ×∠ 360 260c = ° − ° s round a point∠

100c∴ = ° C 90= ° in a semi-circle∠

30d = ° int s of ∠ Δ

90°

90°

130°50° 130°

60°

5

EXERCISE 2 Calculate the value of the unknown variables. O is the centre in each case.

(a) (b) (c)

(d) (e) (f)

(g) (h) (i)

(j) (k) (l) (m) (n) (o)

60°

24°

10°

36°O

B AC

.

120°

x

230°

150°95° 70°

20° 110°

55°

52°

104°

60°

22°

6

EXERCISE 3 (Challenges) O is the centre of each circle.

(a) (b) (c) Prove: Prove: Prove: (1) 1

ˆ ˆO 2C= (1) 1 1ˆˆ2B O 180+ = ° (1) 1

ˆ ˆO 360 2B= ° − (2) ˆ ˆB D 90+ = ° (2) 1

ˆB A 90+ = ° (2) ˆ ˆC 2B 180= − ° THEOREMS INVOLVING CYCLIC QUADRILATERALS Cyclic quadrilaterals A quadrilateral whose vertices lie on the circumference of a circle is referred to as a cyclic quadrilateral. ABCD is a cyclic quadrilateral because A, B, C and D are concyclic (lie on the circumference). THEOREM 4 Angles subtended by a chord (or an arc) of the circle, on the same side of the chord (or the arc), are equal. We say that the angles in the same segment of the circle are equal. ( s in the same seg) In the diagram, 1 1

ˆ ˆA B= , 2 2ˆ ˆA D= , 1 1

ˆ ˆC D= and 2 2ˆB C=

THEOREM 4 (Converse) If a line segment joining two points subtends equal angles at two other points on the same side of the line segment, then these four points are concyclic (lie on the circumference). ABCD will be a cyclic quadrilateral. (line subt = ∠s )

∠⊕

⊕

θθ θθ

7

Corollaries (a) Equal chords subtend equal angles at the circumference.

(equal chords ; equal s∠ )

ˆ ˆA D= (b) Equal chords subtend equal angles at the centre.

(equal chords ; equal s∠ ) ˆ ˆBOC FOE=

(c) Equal chords of equal circles subtend equal angles at the circumference. (equal circles ; equal chords ; equal s∠ ) ˆ ˆA D= and ˆ ˆG H=

EXAMPLE 3

Calculate the value of the unknown angles.

Solutions

60x = ° s∠ in the same segment 130y = ° equal chords, equal angles 60z = ° s∠ opp = sides

EXERCISE 4

Calculate the value of the unknown angles. O is the centre. (a) (b) (c) (d) (e) (f)

18°

22° 16°

35°

32°

92°88°

40°

60°

130°

8

(g) (h) (i) (j) (k) (l)

(m) The two circles have diameters that are equal. O is the centre of circle ABE. EXERCISE 5 (Challenges) (a) (b) (c)

Prove: Prove: Prove: (1) 1

ˆ ˆE B= 1ˆB 90 O= ° − 1

ˆ ˆA 2B 90= − ° (2) ˆˆ ˆBDF A E= + (O is the centre) (O is the centre) (d) In the diagram, chord EF = chord GH. A, B, H, G, F and E are concyclic. Prove that ABCD is a cyclic quadrilateral.

20°32°

43°

110°

20°

15°

35°

120°

110°

9

(e) In the diagram, O and P are the centres of two circles intersecting at C and EC ED= . AB and BC are chords of the larger circle. CD is a chord of the smaller circle. Prove that ABEP is a cyclic quadrilateral. THEOREM 5 The opposite angles of a cyclic quadrilateral are supplementary (add up to ). (opp s of cyclic quad) If AB is a cyclic then the opposite quadrilateral angles are supplementary ˆ ˆB D 180+ = ° and ˆ ˆA C 180+ = ° Proof

In cyclic quadrilateral ABCD, join AO and OC.

1ˆ ˆO 2D= at centre = at circ

2ˆ ˆO 2B= at centre = at circ

1 2ˆ ˆ ˆ ˆO O 2D 2B∴ + = +

and 1 2ˆ ˆO O 360+ = ° s∠ round a point

ˆ ˆ360 2D 2B∴ ° = + ˆ ˆ180 D B∴ ° = +

Similarly, by joining BO and DO, it can be proven that ˆ ˆA C 180+ = ° THEOREM 5 (Converse) If the opposite angles of a quadrilateral are supplementary, then the quadrilateral is a cyclic quadrilateral. (opp s quad supp) If the opposite angles then ABCD is a of quadrilateral ABCD cyclic quadrilateral are supplementary THEOREM 6 An exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. (ext of cyclic quad)

If ABCD is a cyclic then 1ˆ ˆC A=

quadrilateral

180°∠

∠ 2× ∠

∠ 2× ∠

∠

∠

x

180 x° −

x

180 x° −

x

180 x° −

10

THEOREM 6 (Converse) If an exterior angle of a quadrilateral is equal to the interior opposite angle, then the quadrilateral is a cyclic quadrilateral. (ext = int opp )

If 1ˆ ˆC A= then ABCD is a

cyclic quadrilateral EXAMPLE 4 ABCD and BCDE are cyclic quadrilaterals. Calculate the value of the unknown angles. Solutions

80a = ° opp s∠ of cyclic quad ABCD 70b = ° ext ∠ of cyclic quad BCDE 80c = ° s∠ in the same segment

(or opp s∠ of cyclic quad BCDE) 30d = ° s∠ in the same segment 70e = ° int s of ∠ Δ

EXERCISE 6 Calculate the value of the unknown angles. O is the centre. (a) (b) (c) (d) (e) (f)

(g) (h) (i)

∠ ∠

100°30°

70°

105°83°

40°

82°

62°

110°

80°

80°

3y

y

x

3x

y

2x

2y

84°

50°

150°

11

(j) (k) (l)

(m) (n) (o) EXERCISE 7 (Challenges) O is the centre of each circle. (a) (b) (c) Prove: Prove: Prove: (1) 1

ˆ ˆA D 180+ = ° (1) 1 1ˆ ˆO 2Y= (1)

(2) 1 1ˆ ˆO 360 2A= ° − (2) 2

ˆ ˆY K= (2) Important Summary (strategies to prove that a quadrilateral is cyclic) ABCD is a quadrilateral. One of the following three strategies can be used to show that ABCD a cyclic quadrilateral. Strategy 1: Show that at least one pair of opposite angles are supplementary ˆ ˆA C (180 ) 180y y+ = + ° − = ° or ˆ ˆB D (180 ) 180x x+ = + ° − = °

110°

85°

125°

112°

92°

106° 20°

1 1ˆ ˆO 180 2B= ° −

1ˆ ˆO 360 2D= ° −

x

180 x° −

y 180 y° −

12

P

L M

RQ

1 12 2

A

B C

D

E

M1

2

12

E

C

D

A

BF

1

1

2

280°

F DI

CHG

1

1

2

2

Strategy 2: Show that an exterior angle is equal to the interior opposite angle. For example, if 1

ˆ ˆC A= , then ABCD will be a cyclic quadrilateral. Strategy 3: Show that a line segment subtends equal angles on the same side of that line segment. For example, since it is given that line segment AB subtends equal angles ( 1 1

ˆ ˆC D= ), we can conclude that ABCD is a cyclic quadrilateral.

EXERCISE 8 (a) (b)

(c) In the diagram, figure ABCD is a (d) A circle through C and D cuts parallelogram with B 80= ° . parallelogram CDFG at H ABFE is a cyclic quadrilateral. and I. Prove that GFIH is

Show that DEFC is a cyclic cyclic quadrilateral. quadrilateral.

Write down, with reasons, two cyclic quadrilaterals in the figure.

Show that LMRQ is a cyclic quadrilateral if PQ PR= and LM || QR.

13

THEOREMS INVOLVING TANGENTS TO A CIRCLE AXIOM 7 A tangent to a circle is perpendicular to the radius at the point of contact. (tan ⊥ radius)

If ABC is a tangent then OB ABC⊥ to the circle at B AXIOM 7 (Converse) If a line is drawn perpendicular to a radius at the point where the radius meets the circle, then the line is a tangent to the circle. (line ⊥ radius) If line ABC is then ABC is a ⊥ to radius OB tangent at B. THEOREM 8 Two tangents drawn to a circle from the same point outside the circle are equal in length. (tangents from same pt)

THEOREM 9 The angle between a tangent to a circle and a chord drawn from the point of contact is equal to the angle in the alternate segment. ( between tangent and chord) or (tan-chord theorem)

If CDE is a tangent then 1ˆD A=

to the circle at D and 2ˆ ˆC B=

Proof Case 1 (Acute angles) We will prove that ˆ ˆCBD BED= Draw diameter BOF and join EF

1 2ˆ ˆB B 90+ = ° tangent radius

1 2ˆ ˆE E 90+ = ° ∠ in semi-circle

But 1 1ˆ ˆB E= s∠ in the same segment

2 2ˆ ˆB E∴ =

ˆ ˆCBD BED∴ =

∠

⊥

14

Case 2 (Obtuse angles)

We will prove that ˆ ˆABE BDE= Draw diameter BOF and join FD

1B 90= ° tangent radius

1D 90= ° in semi-circle

1 1ˆ ˆB D∴ =

But 2 2ˆ ˆB D= s∠ in the same segment

1 2 1 2ˆ ˆ ˆ ˆB B D D∴ + = +

ˆ ˆABE BDE∴ =

THEOREM 9 (Converse) If a line is drawn through the endpoint of a chord, making with the chord an angle equal to an angle in the alternate segment, then the line is a tangent to the circle. ( between line and chord) or (converse of tan-chord theorem) EXAMPLE 5 Calculate the value of the unknown angles. ABC is a tangent to each circle. (a) Solutions

55a = ° tan ⊥ radius 90b = ° ∠ in semi-circle 1F 65 180c+ + ° = ° int s∠ of Δ BC FC= tangents from same pt 1

ˆF C= s∠ opp = sides 2 66 180c∴ + ° = ° 57c∴ = ° FC is also a tangent O is the centre (b) Solutions 72x = ° tan-chord theorem 65y = ° tan-chord theorem

⊥

∠

∠

65°72°

66°35°

15

(c) Solutions 70x = ° co-int s∠ ; ABC||FG 1B 40= ° alt s∠ ; DE||BF

1By x= + tan-chord theorem

40 70

110y

y∴ = ° + °

= °

EXERCISE 9 Calculate the value of the unknown angles. O is the centre and ABT is a tangent.

(a) (b) (c)

(d) (e) DT is a tangent. DT is a tangent.

(f) (g) (h) (i) (j) (k)

110°

40°

35° 2a

3a50°

80° 60°

60°

70°

40°

100°

30° 100°

35°

7x6x 140°

16

(l) (m) (n) (o) (p) (q) EXERCISE 10 (a) Prove that PT is a tangent to the circle if (b) Prove that SRT is a tangent

4ˆP 60 , BCP 120= ° = ° and 1 2

ˆ ˆC C= to the circle through PQR.

(c) In the diagram, S is the centre of the circle and SQ PR⊥ . TP 12= , PR TU 8= = and SQ 3= . Prove that TP is a tangent to the circle. (d) O is the centre of the circle through A, B, C and T. DTE is a tangent at T and chord BC = chord CT (1) Why is 1 2

ˆ ˆA A= ? (2) Prove that 1 3

ˆ ˆO 2T= (3) Prove that 2 1 1

ˆˆ ˆT B C 180+ + = °

26°

36°32° 40°

56°

40°

59°130°

38°

54°

120°

40°

64°

5x

20x + °

4x

17

(e) In the diagram, DE is a tangent to the circle ACD at D. BC||DE. Prove that CD is a tangent to the circle passing through A, B and C. EXERCISE 11 (NUMERICAL GEOMETRY PROBLEMS) (a) In the diagram below, AOB is a diameter

of the circle centre O. OD||BC. The length of the radius is 10 units.

(1) What is the size of ? State a reason. (2) What is the size of ? State a reason. (3) Why is ? State a reason. (4) If , calculate the length

of ED. (b) In the figure below, RDS is a tangent to the circle centre O at D. ˆBC DC and CDS 40= = ° (1) What is the size of . State a reason. (2) What is the size of . State a reason.

(3) What is the size of . State a reason. (4) Calculate the size of State a reason. (5) Calculate the size of . State a reason. (6) Calculate the size of State reasons. (7) Calculate the size of State a reason.

(c) In the diagram, O is the centre

of the circle passing through A, B, C and D. AB||CD and

(1) Calculate the size of ? State a reason. (2) Calculate the size of ? State a reason. (3) Calculate the size of ? State a reason. (4) Calculate the size of ? State a reason. (5) Why is AOEC a cyclic quadrilateral?

C1E

AE EC=AC 16 units=

1B

2D

C2O

1O

3DA

B 20= °

1C

1OD

1E

> >40°

20°

18

D

E

A

F

O

BC

A

B

D

O

C

21

112 2 3

E

C

A

D

OB21

1

21

2 3

B

A

T

O U

2

1

121

2 3

S

C

3

11 2

(d) AB and CD are two chords of the circle centre O. OE ⊥ CD, AF FB= . OE 4 cm, OF 3 cm= = and AB 8 cm= . Calculate the length of CD. (e) In the circle centre O, BO OD⊥ ,

AE EC= and ˆAOD 116= ° . Calculate the size of: (1) C (2) ˆABC

(f) O is the centre of the circle.

AB is the diameter. ˆAOC 104= ° and ˆDAB 32= ° .

Calculate the size of: (1) D (2) 3C (3) 1A (4) 2C

(g) O is the centre of the circle.

STU is a tangent at T. Chord BC = chord CT,

ˆATC 105= ° and ˆCTU 40= ° . Calculate the size of : (1) 2A (2) 1A (3) 1 2

ˆ ˆB B+ (4) 2C

19

Q

S

V

W

2

1

1

2

2 3

T

R1

1 2

30°

70°

P

Q

P

T

O

2

1

1 2

S

R

12

1

2

21

35°

(h) PQR is a tangent at Q. ST || QW.

ˆWQR 30= ° and ˆTSW 70= ° . Calculate the size of the following angles with reasons: (1) V (2) 1Q (3) 1T (4) 2W

(i) O is the centre of circle ABC.

AC is produced to D so that CD = CB. Prove that 1O 128= ° if D 32= ° .

(j) QOS is a diameter of the circle centre O. QR = RT and T 35= ° .

Calculate, with reasons, the size of: (1) 2R (2) 2S (3) 2Q (4) 1Q (5) 1P

(k) In the diagram, O is the centre of the circle and AO||BC.

If , calculate the size of .

(l) In the diagram below, JM||KL and the centre of the circle is O. .

(1) Calculate the size of . (2) Prove that XYLK is a cyclic

quadrilateral.

1A 130= ° 1O

1O 120= °

1Y

130°

120°

20

(GEOMETRY PROBLEMS INVOLVING VARIABLES) (m) In the diagram, ABCD is a cyclic quadrilateral with AB parallel to chord ED. 1A x= and AB bisects ˆNAG . TAN is a tangent to the circle at A. (1) Write down, with reasons, seven angles equal to x. (2) Why is AB a tangent to the circle passing through A, G and E?

(n) In the diagram, chord PR is parallel to chord TS. PR PS= and PT TS= . ARB is a tangent to the circle at R and RST is a straight line. 3R x= and 1R y= . (1) Write down, with reasons, three other angles equal to x and three other equal to y. (2) Express T in terms of y. (3) Express T in terms of x. (o) In the diagram, DE is a tangent to circle ABCD at D. AD||BC and 1 3

ˆ ˆB B= .

(1) Why is 3D x= ? (2) Why is DEFB a cyclic quadrilateral? (3) Why is 5B x= ?

(4) Why is 1D x= ? (Don’t assume that EB is a tangent!) (5) Why can we now conclude that EB is a tangent at B? (p) In the diagram, O is the centre of the circle and E is the midpoint of chord AD. AO and EO are produced to cut the circle at C and B respectively. AD produced meets the tangent drawn to the circle at C in G. 3O x= (1) Why is EOCG is a cyclic quadrilateral? (2) Write down, with reasons, two other angles equal to x. (3) Express 1A in terms of x.

(4) Express 1C in terms of x.

(5) Express 2C in terms of x.

A

P

T

B

R

1

332

x

2

12

1

S

y

=

=

A4

1

E

D

B

C

21

332

Fx1 2

1 2

1 2

T

N

G

A

1

E

D

BC

2

1

3

32

x 12

12

O

G

4

1 2 =

=

>

>

21

A

D

C

B

O

32

1

21

1 2

2x

(q) O is the centre of the circle through A, B, C and D. BC = CD and ˆBOD 2x= . Express the following in terms of x. (1) 2B (2) ˆBCD (3) A

(r) In the diagram, AB is a diameter of the circle. EA is a tangent to the circle at A. 1B x= and E y= . (1) Name one other angle equal to x. (2) Show that 2A y= . (3) Name one other angle equal to y. (s) In the diagram, AOB is the diameter of the semi-circle AMNB. MO||NB and 1B x= .

(1) Express 2M and 2B in terms of x.

(2) Express 2O and 2N in terms of x.

(3) Express 1K in terms of x. *(4) If it is given that 30x = ° , calculate the sizes of the angles of MKNΔ . *(5) Hence prove that MOBN is a rhombus. SOLVING GEOMETRICAL RIDERS Some important concepts A rider is a geometrical problem requiring proofs of statements. The problem involves the application of a combination of theorems and logical abstract reasoning. The following statements of logic are extremely important for solving geometrical riders. (a) If a b= (b) If a b c+ = (c) If a b= and b c= and a d c+ = and c d= then a c= then b d= then a c b d+ = + (d) If a b c d+ = + (e) If a b c d e+ + = + (f) If a b c d+ + = and b c= and b d= and a b= then a d= then a c e+ = then 2b c d+ = or 2a c d+ =

A1

M

K

B

213

3

x

12

2

O

1 2

N

1

22

EXAMPLE 1 CD is a tangent at C to the circle through A, B, C and E. BC CE= and chords AC and BE intersect at F. Prove that: (a) 1 2

ˆ ˆC A=

(b) 4 2ˆ ˆC E=

Solutions

(a) 1 1ˆ ˆC A= tan-chord theorem

1 2ˆ ˆA A= equal chords ; equal angles

1 2ˆ ˆC A∴ =

(b) 4 2ˆ ˆC A= tan-chord theorem

2 1ˆ ˆA A= equal chords ; equal angles

4 1ˆ ˆC A∴ =

1 2ˆ ˆA E= s∠ in the same segment

4 2ˆ ˆC E∴ =

EXAMPLE 2

PR is a diameter of circle PRMS with centre Q. PS, SR and PM are chords. PM bisects ˆRPS. Prove that: (a) PS||QM (b) QM SR⊥ (c) QM bisects SR Solutions

(a) 2ˆ ˆP M= s∠ opp = radii

2 1ˆ ˆP P= given

1ˆP M∴ =

∴ PS||QM alt s∠ = (b) S 90= ° in semi-circle∠ 2E 90∴ = ° corr s∠ = QM SR∴ ⊥ (c) RE ES= line from centre ⊥ to chord ∴QM bisects SR

Alternative for (b)

4 2ˆ ˆC A= tan-chord theorem

2 2ˆ ˆA B= s∠ in the same segment

4 2ˆ ˆC B∴ =

2 2ˆ ˆB E= s∠ opp = sides

4 2ˆ ˆC E∴ =

23

EXAMPLE 3 LOM is a diameter of circle LMT. The centre is O. TN is a tangent at T. LN NP⊥ . MT is a chord. LT is a chord produced to P. Prove that: (a) MNPT is a cyclic quadrilateral (b) NP = NT Solutions

(a) 1 2ˆ ˆN N 90+ = ° given

3T 90= ° in semi-circle∠

1 2 3ˆ ˆ ˆN N T∴ + =

∴MNPT is a cyclic quad ext ∠ of quad = int opp ∠

(b) 1 4ˆ ˆT T= vert opp s∠ =

4 1ˆ ˆT M= tan-chord theorem

1 1ˆ ˆT M∴ =

1ˆ ˆM P= ext ∠ of cyclic quad

1ˆ ˆT P∴ =

NP NT∴ = sides opp s= ∠ EXAMPLE 4 AB, AC, DB, and DC are chords. DE AC⊥ and DB FC⊥ . Prove that: (a) DEFC is a cyclic quadrilateral (b) AB || EF Solutions

(a) 1E 90= ° given

2F 90= ° given ∴ DEFC is a cyclic quad DC subtends s= ∠

(b) 2 2ˆ ˆE D= s∠ in the same seg

2ˆ ˆA D= s∠ in the same seg

2ˆE A∴ =

∴AB || EF corr s∠ =

24

1

1

12

2

2

A Q

ST

P

R

O1 2

E

EXAMPLE 5 O is the centre of circle SAT which is inscribed in PQRΔ . PQ, QR and PR are tangents to the circle. Prove: (a) PSOT is a cyclic quadrilateral (b) OS is a tangent to circle SPE Solutions

(a) 1 2ˆ ˆT T 90+ = ° tan ⊥ radius

1 2ˆ ˆS S 90+ = ° tan ⊥ radius

1 2 1 2ˆ ˆˆ ˆT T S S 180∴ + + + = °

∴PSOT is a cyclic quadrilateral opp s∠ of quad supp

(b) 1 1ˆ ˆS T= s∠ opp = radii

2 1ˆ ˆP T= s∠ in the same seg

1 2ˆ ˆS P∴ =

∴OS is a tangent to circle SPE ∠ between line and chord EXAMPLE 6

A, D, C, B and P are concyclic. PC bisects ˆDCB. Prove that PA bisects ˆXAB.

Solution

1 1ˆ ˆA C= ext ∠ of cyclic quad

1 2ˆ ˆC C= given

1 2ˆ ˆA C∴ =

2 2ˆ ˆC A= s∠ in the same seg

1 2ˆ ˆA A∴ =

25

2 1

A

P

B

O C

1

B

P

R

Q2

S..

1D

AB

E

2 C

21

31 2

21..

1A

ED

B

2

C

21

3

E

A

B C2

D

31

21

EXERCISE 12 (SOLVING RIDERS) (a) ABC is a tangent to the circle BED.

BE || CD. Prove that 1

ˆD C=

(c) AC is a diameter of circle centre O.

B is a point on the circle. OP bisects AB. Prove that OP || BC.

(e) ABCD is a cyclic quadrilateral. BA is produced to E. AD bisects ˆEAC . Prove that DC = DB.

1

R

Q

B

A

2

P

21

(d) PQB is a tangent to the circle QRS at Q. QS bisects ˆBQR . Prove that QS = RS

(f) AQ is a tangent to the circle in Q and QP || AR. Prove that RA is a tangent to circle ABQ at A.

(b) In circle ABCDE, BC and AD are parallel chords. AB is produced to F. (1) Name 2 cyclic quadrilaterals (2) Prove that: (i) 1

ˆ ˆB E=

(ii) 1ˆD A=

26

C

B

AD

E

1 2

12

3

21

21

21

421

B

QP

N

M

LA

3

21

2 1

4

3

C

BA

D

E

122

1

O.F

B

D

A

C

P

21

4 321

43 2 1

21

43

21

(g) In circle ABCD, AB = BC. Prove that AB is a tangent to circle AED in A.

(i) EC is a diameter of circle DEC.

EC is produced to B. BD is a tangent at D. ED is produced to A and AB BE⊥ . Prove that: (1) ABCD is a cyclic quadrilateral (2) 1

ˆ ˆA E= (3) BDAΔ is isosceles (4) 2 3

ˆ ˆC C=

(h) ALB is a tangent to circle LMNP. ALB || MP. Prove that:

(1) LM = LP (2) LN bisects ˆMNP (3) LM is a tangent to circle MNQ

(j) PA and PC are tangents to the circle at A and C. AD || PC, and PD cuts the circle B. CB is produced to meet AP at F. AB, AC and DC are drawn. Prove: (1) AC bisects ˆPAD (2) 1 3

ˆ ˆB B= (3) ˆ ˆAPC ABD=

27

121

21

1

32

1

P

R

AT

MO

(k) TA is a tangent to the circle PRT. M is the midpoint of chord PT. The centre of the circle is O. PR is produced to intersect TA at A and TA PA⊥ . T and R are joined. OR and OT are radii. Prove that: (1) MTAR is a cyclic quadrilateral (2) PR = RT (3) TR bisects ˆPTA

(4) 2 11 ˆT O2

=

EXAMPLE 6 (MORE ADVANCED RIDERS) In the diagram, PA is a tangent to the circle at A. AC and AB are chords and PM is produced to K such that AK AM= . K and M lie on AC and AB respectively. Chord CB is produced to P. Prove that KP bisects ˆAPC . Solution

1 1 1ˆˆ ˆP A M+ = ext ∠ of AMPΔ

1 1ˆ ˆBut M K= s opp sides∠ =

1 1 1ˆˆ ˆP A K∴ + =

2 1ˆˆ ˆBut P C K+ = ext ∠ of CKPΔ

1ˆ ˆand A C= tan-chord theorem

1 2ˆ ˆP P∴ =

EXERCISE 13

(a) FC is a tangent to circle BCD and FG is a tangent to circle FEDC. The circles intersect at D and C. Chord BD is produced to F. DC is joined. Chord FE and chord CB produced meet at A. Prove that: (1) ABDE is a cyclic quadrilateral. (2) 1 1 2

ˆ ˆD C C= + (3) FG||AC

A

B

P

C

M

2

2

11

1

K

1

12

==

A

B

D

F

G

2

21

1

1E

1

12

C

28

OE

B

F

C

D

2

1

11

2

(b) LPN is a tangent to circle ADP. Circle BCP touches the larger circle internally at P. Chord AD cuts the smaller circle at B and C and BP and CP are joined. Prove that 2 4

ˆ ˆP P= CONSOLIDATION AND EXTENSION EXERCISE (a) In the diagram below, QP is a tangent

to a circle with centre O. RS is a diameter of the circle and RQ is a straight line. T is a point on the circle. PS bisects ˆTPQ and ˆSPQ 22= ° . Calculate the following, giving reasons: (1) 2P (2) 2R

(3) 3 4ˆ ˆP P+ (4) 1R

(5) 1O (6) 2Q

(b) O is the centre of the circle ABED.

The radius of the circle is 6,5 cm. AC is a tangent to the circle at A. BD is produced to C such that AD DC= . (1) Prove that AB AC= .

(2) If BE 5 cm= , calculate the length of AC.

A

B

P

C

2

1

11

D

L

N

3

4 5

P

21

1

1

R

34

.

T

Q

O

S

2

2

32

1

12

1 2

34

(c) O is the centre of the larger circle. The second circle drawn through O intersects the larger circle at B and D. Chords BC and DC intersect the smaller circle at E and F respectively. Prove that DE = EC.

B C

E

A

O

D

12

.3

12

1 2

29

(c) In the diagram below, 1ˆ ˆA C= ,

B 3 10x= + ° and D 2 20x= + ° .

Calculate, with reasons, the value of x.

(d) In the figure, O is the centre of the circle. PRO is a straight line intersecting the circle at R. OT is perpendicular to chord SQ at T. SR is joined. Radius OQ and line PQ meet at Q. PQ 12 units, QS 8 units, = = RP 8 units and OT 3 = = units. (1) Prove that PQ is a

tangent to the circle at Q.

(2) Express P in terms of x if 4Q = x .

(e) AC is a diameter of the circle

centre B. FED is a tangent to the circle at E and BG EC.⊥ BG produced cuts FE

produced at D. DC is drawn. Prove that:

(1) BG||AE (2) BCDE is a cyclic quadrilateral. (3) DC is a tangent to circle EAC. (4) DC is a tangent to circle BCG.

(f) Circles OAS and ABCS intersect at A and S. SB bisects ˆABC . Chord AT is produced to C.

Prove that SA SO=

A1

B C

D

3 10x + °

2 20x + °

R

Q

S

O

P1

2

.31

2

1

2 34

1 2

T

CB

E

D

A

1

2 31

21

23

41 2

F

34

G

CB

O

A

1

2

1

1

12

1

2

S

T

C

B

Q

A1

2

2

1

1

2

1 2P

R

12

34

(g) P is a point on side AB of ABCΔ . The circle through P, B and C cuts AC in Q. QP produced cuts the larger circle at R. Prove that 1 1 1

ˆˆ ˆP A B= +

30

(h) Two circles with centre O and P touch each other. The radii of the circles are 7 and 4 respectively. Calculate the length of AB if AB is a tangent to both circles. Round off your answer to one decimal place. (i) BC is a diameter of circle ABC with centre O. OD AC⊥ and

AT BOC.⊥ BT x= , TO 5= and AD 42= . (1) Prove that OA 7= (2) Calculate the length of AB. (j) PR is a common chord of circles

PQSR and PXRY. PR bisects ˆXRY . T lies on chord PS such

that ST SR SQ= = . MPY touches the larger circle at P. Prove that: (1) 1 2

ˆ ˆQ Q= (2) QP is a tangent

to the smaller circle. (k) In the figure, ABCD is a cyclic quadrilateral with AB AD= and DC BC= . DC and BC, both produced meet AB and AD, both produced, at E and F respectively. AC produced meets FE at G with 1G 90= ° Prove that: (1) AC is a diameter of the circle. (2) DBEF is a cyclic quadrilateral. (3) BC bisects ˆDBG .

Q

S

M

1

2

2

1

12

1

2

P

Y

1

2

3

R

T

3 4

3X

5

D

A

B

12

1 12

2

3O CT

x

D

A

B

C

F

G

E

H

12 3

4

12

12

3

12

12

12

3

12

12

A B

OP74

..