The document discusses several concepts in circle geometry and trigonometry including: (1) Converse circle theorems such as if two points subtend the same angle from an interval, then they are concyclic. (2) The four centers of a triangle - incentre, circumcentre, centroid, and orthocentre which are the points of concurrency of angle bisectors, perpendicular bisectors, medians, and altitudes respectively. (3) Relationships between trigonometry and geometry such as the circumradius, inradius, and sine rule.

1. Harder Extension 1
Circle Geometry

2. Harder Extension 1
Converse Circle Theorems
Circle Geometry

3. Harder Extension 1
Converse Circle Theorems
Circle Geometry
(1) The circle whose diameter is the hypotenuse of a right angled
triangle passes through the third vertex.

4. Harder Extension 1
Converse Circle Theorems
Circle Geometry
(1) The circle whose diameter is the hypotenuse of a right angled
triangle passes through the third vertex.
A B
C

5. Harder Extension 1
Converse Circle Theorems
Circle Geometry
(1) The circle whose diameter is the hypotenuse of a right angled
triangle passes through the third vertex.
A B
C
ABC are concyclic with AB diameter

6. Harder Extension 1
Converse Circle Theorems
Circle Geometry
(1) The circle whose diameter is the hypotenuse of a right angled
triangle passes through the third vertex.
A B
C
ABC are concyclic with AB diameter
 
90semicircleain 

7. Harder Extension 1
Converse Circle Theorems
Circle Geometry
(1) The circle whose diameter is the hypotenuse of a right angled
triangle passes through the third vertex.
A B
C
ABC are concyclic with AB diameter
 
90semicircleain 
(2) If an interval AB subtends the same angle at two points P and Q on
the same side of AB, then A,B,P,Q are concyclic.

8. Harder Extension 1
Converse Circle Theorems
Circle Geometry
(1) The circle whose diameter is the hypotenuse of a right angled
triangle passes through the third vertex.
A B
C
ABC are concyclic with AB diameter
 
90semicircleain 
(2) If an interval AB subtends the same angle at two points P and Q on
the same side of AB, then A,B,P,Q are concyclic.
A B
P Q
 

9. Harder Extension 1
Converse Circle Theorems
Circle Geometry
(1) The circle whose diameter is the hypotenuse of a right angled
triangle passes through the third vertex.
A B
C
ABC are concyclic with AB diameter
 
90semicircleain 
(2) If an interval AB subtends the same angle at two points P and Q on
the same side of AB, then A,B,P,Q are concyclic.
A B
P Q
  ABQP is a cyclic quadrilateral

10. Harder Extension 1
Converse Circle Theorems
Circle Geometry
(1) The circle whose diameter is the hypotenuse of a right angled
triangle passes through the third vertex.
A B
C
ABC are concyclic with AB diameter
 
90semicircleain 
(2) If an interval AB subtends the same angle at two points P and Q on
the same side of AB, then A,B,P,Q are concyclic.
A B
P Q
  ABQP is a cyclic quadrilateral
 aresegmentsameins 

11. (3) If a pair of opposite angles in a quadrilateral are supplementary (or
if an exterior angle equals the opposite interior angle) then the
quadrilateral is cyclic.

12. (3) If a pair of opposite angles in a quadrilateral are supplementary (or
if an exterior angle equals the opposite interior angle) then the
quadrilateral is cyclic.
The Four Centres Of A Triangle

13. (3) If a pair of opposite angles in a quadrilateral are supplementary (or
if an exterior angle equals the opposite interior angle) then the
quadrilateral is cyclic.
The Four Centres Of A Triangle
(1) The angle bisectors of the vertices are concurrent at the incentre
which is the centre of the incircle, tangent to all three sides.

14. (3) If a pair of opposite angles in a quadrilateral are supplementary (or
if an exterior angle equals the opposite interior angle) then the
quadrilateral is cyclic.
The Four Centres Of A Triangle
(1) The angle bisectors of the vertices are concurrent at the incentre
which is the centre of the incircle, tangent to all three sides.

15. (3) If a pair of opposite angles in a quadrilateral are supplementary (or
if an exterior angle equals the opposite interior angle) then the
quadrilateral is cyclic.
The Four Centres Of A Triangle
(1) The angle bisectors of the vertices are concurrent at the incentre
which is the centre of the incircle, tangent to all three sides.





incentre

16. (3) If a pair of opposite angles in a quadrilateral are supplementary (or
if an exterior angle equals the opposite interior angle) then the
quadrilateral is cyclic.
The Four Centres Of A Triangle
(1) The angle bisectors of the vertices are concurrent at the incentre
which is the centre of the incircle, tangent to all three sides.





incentre incircle

17. (2) The perpendicular bisectors of the sides are concurrent at the
circumcentre which is the centre of the circumcircle, passing
through all three vertices.

18. (2) The perpendicular bisectors of the sides are concurrent at the
circumcentre which is the centre of the circumcircle, passing
through all three vertices.

19. (2) The perpendicular bisectors of the sides are concurrent at the
circumcentre which is the centre of the circumcircle, passing
through all three vertices.
circumcentre

20. (2) The perpendicular bisectors of the sides are concurrent at the
circumcentre which is the centre of the circumcircle, passing
through all three vertices.
circumcentre
circumcircle

21. (2) The perpendicular bisectors of the sides are concurrent at the
circumcentre which is the centre of the circumcircle, passing
through all three vertices.
circumcentre
circumcircle
(3) The medians are concurrent at the centroid, and the centroid trisects
each median.

22. (2) The perpendicular bisectors of the sides are concurrent at the
circumcentre which is the centre of the circumcircle, passing
through all three vertices.
circumcentre
circumcircle
(3) The medians are concurrent at the centroid, and the centroid trisects
each median.

23. (2) The perpendicular bisectors of the sides are concurrent at the
circumcentre which is the centre of the circumcircle, passing
through all three vertices.
circumcentre
circumcircle
(3) The medians are concurrent at the centroid, and the centroid trisects
each median.
centroid

24. (4) The altitudes are concurrent at the orthocentre.

25. (4) The altitudes are concurrent at the orthocentre.

26. (4) The altitudes are concurrent at the orthocentre.
orthocentre

27. (4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry

28. (4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
lecircumcircifdiameter
sinsinsin

C
c
B
b
A
a

29. (4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
lecircumcircifdiameter
sinsinsin

C
c
B
b
A
a
Proof: A
B
C

30. (4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
lecircumcircifdiameter
sinsinsin

C
c
B
b
A
a
Proof: A
B
C
O
P

31. (4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
lecircumcircifdiameter
sinsinsin

C
c
B
b
A
a
Proof: A
B
C
O
P
PA    segmentsamein

32. (4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
lecircumcircifdiameter
sinsinsin

C
c
B
b
A
a
Proof: A
B
C
O
P
PA    segmentsamein
PA sinsin 

33. (4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
lecircumcircifdiameter
sinsinsin

C
c
B
b
A
a
Proof: A
B
C
O
P
PA    segmentsamein
PA sinsin 

90PBC

34. (4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
lecircumcircifdiameter
sinsinsin

C
c
B
b
A
a
Proof: A
B
C
O
P
PA    segmentsamein
PA sinsin 

90PBC  
90semicirclein 

35. (4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
lecircumcircifdiameter
sinsinsin

C
c
B
b
A
a
Proof: A
B
C
O
P
PA    segmentsamein
PA sinsin 

90PBC  
90semicirclein 
P
PC
BC
sin

36. (4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
lecircumcircifdiameter
sinsinsin

C
c
B
b
A
a
Proof: A
B
C
O
P
PA    segmentsamein
PA sinsin 

90PBC  
90semicirclein 
P
PC
BC
sin
PC
P
BC

sin

37. (4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
lecircumcircifdiameter
sinsinsin

C
c
B
b
A
a
Proof: A
B
C
O
P
PA    segmentsamein
PA sinsin 

90PBC  
90semicirclein 
P
PC
BC
sin
PC
P
BC

sin A
BC
PC
sin


38. e.g. (1990)
In the diagram, AB is a fixed chord of a circle, P a variable point in the
circle and AC and BD are perpendicular to BP and AP respectively.

39. e.g. (1990)
In the diagram, AB is a fixed chord of a circle, P a variable point in the
circle and AC and BD are perpendicular to BP and AP respectively.
(i) Show that ABCD is a cyclic quadrilateral on a circle with AB as
diameter.

40. e.g. (1990)
In the diagram, AB is a fixed chord of a circle, P a variable point in the
circle and AC and BD are perpendicular to BP and AP respectively.
(i) Show that ABCD is a cyclic quadrilateral on a circle with AB as
diameter.

90 ACBBDA

41. e.g. (1990)
In the diagram, AB is a fixed chord of a circle, P a variable point in the
circle and AC and BD are perpendicular to BP and AP respectively.
(i) Show that ABCD is a cyclic quadrilateral on a circle with AB as
diameter.

90 ACBBDA  given

42. e.g. (1990)
In the diagram, AB is a fixed chord of a circle, P a variable point in the
circle and AC and BD are perpendicular to BP and AP respectively.
(i) Show that ABCD is a cyclic quadrilateral on a circle with AB as
diameter.

90 ACBBDA  given
ralquadrilatecyclicaisABCD

43. e.g. (1990)
In the diagram, AB is a fixed chord of a circle, P a variable point in the
circle and AC and BD are perpendicular to BP and AP respectively.
(i) Show that ABCD is a cyclic quadrilateral on a circle with AB as
diameter.

90 ACBBDA  given
ralquadrilatecyclicaisABCD   aresegmentsameins

44. e.g. (1990)
In the diagram, AB is a fixed chord of a circle, P a variable point in the
circle and AC and BD are perpendicular to BP and AP respectively.
(i) Show that ABCD is a cyclic quadrilateral on a circle with AB as
diameter.

90 ACBBDA  given
ralquadrilatecyclicaisABCD   aresegmentsameins

90semicircleinasdiameteris AB

45. (ii) Show that triangles PCD and APB are similar

46. (ii) Show that triangles PCD and APB are similar
DPCAPB 

47. (ii) Show that triangles PCD and APB are similar
DPCAPB   scommon 

48. (ii) Show that triangles PCD and APB are similar
DPCAPB   scommon 
PBAPDC 

49. (ii) Show that triangles PCD and APB are similar
DPCAPB   scommon 
PBAPDC   ralquadrilatecyclicexterior

50. (ii) Show that triangles PCD and APB are similar
DPCAPB   scommon 
PBAPDC   ralquadrilatecyclicexterior
PBAPDC  |||

51. (ii) Show that triangles PCD and APB are similar
DPCAPB   scommon 
PBAPDC   ralquadrilatecyclicexterior
PBAPDC  |||  requiangula

52. (iii) Show that as P varies, the segment CD has constant length.

53. (iii) Show that as P varies, the segment CD has constant length.
P
A B
P
C D

54. (iii) Show that as P varies, the segment CD has constant length.
AP
PC
AB
CD

P
A B
P
C D

55. (iii) Show that as P varies, the segment CD has constant length.
AP
PC
AB
CD
  s|||insidesofratio 
P
A B
P
C D

56. (iii) Show that as P varies, the segment CD has constant length.
AP
PC
AB
CD
  s|||insidesofratio 
P
AP
PC
PCA cos,In 
P
A B
P
C D

57. (iii) Show that as P varies, the segment CD has constant length.
AP
PC
AB
CD
  s|||insidesofratio 
P
AP
PC
PCA cos,In 
P
AB
CD
cos
P
A B
P
C D

58. (iii) Show that as P varies, the segment CD has constant length.
AP
PC
AB
CD
  s|||insidesofratio 
P
AP
PC
PCA cos,In 
P
AB
CD
cos
PABCD cos
P
A B
P
C D

59. (iii) Show that as P varies, the segment CD has constant length.
AP
PC
AB
CD
  s|||insidesofratio 
P
AP
PC
PCA cos,In 
P
AB
CD
cos
PABCD cos
constantisNow, P
P
A B
P
C D

60. (iii) Show that as P varies, the segment CD has constant length.
AP
PC
AB
CD
  s|||insidesofratio 
P
AP
PC
PCA cos,In 
P
AB
CD
cos
PABCD cos
constantisNow, P
P
A B
P
C D
  aresegmentsameins

61. (iii) Show that as P varies, the segment CD has constant length.
AP
PC
AB
CD
  s|||insidesofratio 
P
AP
PC
PCA cos,In 
P
AB
CD
cos
PABCD cos
constantisNow, P
P
A B
P
C D
fixedisand AB
  aresegmentsameins

62. (iii) Show that as P varies, the segment CD has constant length.
AP
PC
AB
CD
  s|||insidesofratio 
P
AP
PC
PCA cos,In 
P
AB
CD
cos
PABCD cos
constantisNow, P
P
A B
P
C D
fixedisand AB
  aresegmentsameins
 given

63. (iii) Show that as P varies, the segment CD has constant length.
AP
PC
AB
CD
  s|||insidesofratio 
P
AP
PC
PCA cos,In 
P
AB
CD
cos
PABCD cos
constantisNow, P
P
A B
P
C D
fixedisand AB
  aresegmentsameins
 given
constantisCD

64. (iv) Find the locus of the midpoint of CD.

65. (iv) Find the locus of the midpoint of CD.
ABCD is a cyclic quadrilateral with AB diameter.
A B
CD

66. (iv) Find the locus of the midpoint of CD.
ABCD is a cyclic quadrilateral with AB diameter.
A B
CD Let M be the midpoint of CD
M

67. (iv) Find the locus of the midpoint of CD.
ABCD is a cyclic quadrilateral with AB diameter.
A B
CD Let M be the midpoint of CD
O is the midpoint of AB
M
O

68. (iv) Find the locus of the midpoint of CD.
ABCD is a cyclic quadrilateral with AB diameter.
A B
CD Let M be the midpoint of CD
O is the midpoint of AB
M
O
OM is constant

69. (iv) Find the locus of the midpoint of CD.
ABCD is a cyclic quadrilateral with AB diameter.
A B
CD Let M be the midpoint of CD
O is the midpoint of AB
M
O
OM is constant
 centrethefromtequidistanarechords

70. (iv) Find the locus of the midpoint of CD.
ABCD is a cyclic quadrilateral with AB diameter.
A B
CD Let M be the midpoint of CD
O is the midpoint of AB
M
O
OM is constant
 centrethefromtequidistanarechords
OM fromdistancefixedais

71. (iv) Find the locus of the midpoint of CD.
ABCD is a cyclic quadrilateral with AB diameter.
A B
CD Let M be the midpoint of CD
O is the midpoint of AB
M
O
OM is constant
 centrethefromtequidistanarechords
OM fromdistancefixedais
222
MCOCOM 

72. (iv) Find the locus of the midpoint of CD.
ABCD is a cyclic quadrilateral with AB diameter.
A B
CD Let M be the midpoint of CD
O is the midpoint of AB
M
O
OM is constant
 centrethefromtequidistanarechords
OM fromdistancefixedais
222
MCOCOM 
PAB
PABAB
PABAB
22
222
22
sin
4
1
cos
4
1
4
1
cos
2
1
2
1
















73. (iv) Find the locus of the midpoint of CD.
ABCD is a cyclic quadrilateral with AB diameter.
A B
CD Let M be the midpoint of CD
O is the midpoint of AB
M
O
OM is constant
 centrethefromtequidistanarechords
OM fromdistancefixedais
222
MCOCOM 
PAB
PABAB
PABAB
22
222
22
sin
4
1
cos
4
1
4
1
cos
2
1
2
1















PABOM sin
2
1


74. (iv) Find the locus of the midpoint of CD.
ABCD is a cyclic quadrilateral with AB diameter.
A B
CD Let M be the midpoint of CD
O is the midpoint of AB
M
O
OM is constant
 centrethefromtequidistanarechords
OM fromdistancefixedais
222
MCOCOM 
PAB
PABAB
PABAB
22
222
22
sin
4
1
cos
4
1
4
1
cos
2
1
2
1















PABOM sin
2
1

PAB
O
sin
2
1
radiusand
centrecircle,islocus



75. 2008 Extension 2 Question 7b)
In the diagram, the points P, Q and R lie on a circle. The tangent at P
and the secant QR intersect at T. The bisector of meets QR at S
so that . The intervals RS, SQ and PT have lengths
a, b and c respectively.
PQR
QPS RPS    

76. 2008 Extension 2 Question 7b)
In the diagram, the points P, Q and R lie on a circle. The tangent at P
and the secant QR intersect at T. The bisector of meets QR at S
so that . The intervals RS, SQ and PT have lengths
a, b and c respectively.
PQR
QPS RPS    
( ) Show thati TSP TPS  

77. 2008 Extension 2 Question 7b)
In the diagram, the points P, Q and R lie on a circle. The tangent at P
and the secant QR intersect at T. The bisector of meets QR at S
so that . The intervals RS, SQ and PT have lengths
a, b and c respectively.
PQR
QPS RPS    
( ) Show thati TSP TPS  
RQP RPT  

78. 2008 Extension 2 Question 7b)
In the diagram, the points P, Q and R lie on a circle. The tangent at P
and the secant QR intersect at T. The bisector of meets QR at S
so that . The intervals RS, SQ and PT have lengths
a, b and c respectively.
PQR
QPS RPS    
( ) Show thati TSP TPS  
RQP RPT    alternate segment theorem

79. 2008 Extension 2 Question 7b)
In the diagram, the points P, Q and R lie on a circle. The tangent at P
and the secant QR intersect at T. The bisector of meets QR at S
so that . The intervals RS, SQ and PT have lengths
a, b and c respectively.
PQR
QPS RPS    
( ) Show thati TSP TPS  
RQP RPT    alternate segment theorem
TSP RQP SPQ    

80. 2008 Extension 2 Question 7b)
In the diagram, the points P, Q and R lie on a circle. The tangent at P
and the secant QR intersect at T. The bisector of meets QR at S
so that . The intervals RS, SQ and PT have lengths
a, b and c respectively.
PQR
QPS RPS    
( ) Show thati TSP TPS  
RQP RPT    alternate segment theorem
TSP RQP SPQ      exterior , SPQ 

81. 2008 Extension 2 Question 7b)
In the diagram, the points P, Q and R lie on a circle. The tangent at P
and the secant QR intersect at T. The bisector of meets QR at S
so that . The intervals RS, SQ and PT have lengths
a, b and c respectively.
PQR
QPS RPS    
( ) Show thati TSP TPS  
RQP RPT    alternate segment theorem
TSP RQP SPQ      exterior , SPQ 
TSP RPT    

82. SPT RPT    

83. SPT RPT      common 

84. SPT RPT      common 
SPT TSP  

85. SPT RPT      common 
SPT TSP  
1 1 1
( ) Hence show thatii
a b c
 

86. SPT RPT      common 
SPT TSP  
1 1 1
( ) Hence show thatii
a b c
 

87. SPT RPT      common 
SPT TSP  
1 1 1
( ) Hence show thatii
a b c
 
is isoscelesTPS

88. SPT RPT    
 common 
 2 = 's
SPT TSP  
1 1 1
( ) Hence show thatii
a b c
 
is isoscelesTPS
SPT RPT    

89. SPT RPT    
SPT TSP  
1 1 1
( ) Hence show thatii
a b c
 
is isoscelesTPS  2 = 's
 common SPT RPT    

90. SPT RPT    
SPT TSP  
1 1 1
( ) Hence show thatii
a b c
 
is isoscelesTPS
ST c 
 2 = 's
 common SPT RPT    

91. SPT RPT    
SPT TSP  
1 1 1
( ) Hence show thatii
a b c
 
is isoscelesTPS
ST c   = sides in isosceles 
 2 = 's
 common SPT RPT    

92. SPT RPT    
SPT TSP  
1 1 1
( ) Hence show thatii
a b c
 
is isoscelesTPS
ST c   = sides in isosceles 
2
PT QT RT 
 2 = 's
 common SPT RPT    

93. SPT RPT    
SPT TSP  
1 1 1
( ) Hence show thatii
a b c
 
is isoscelesTPS
ST c   = sides in isosceles 
2
PT QT RT 
 square of tangents=products of intercepts
 2 = 's
 common SPT RPT    

94. SPT RPT    
SPT TSP  
1 1 1
( ) Hence show thatii
a b c
 
is isoscelesTPS
ST c   = sides in isosceles 
2
PT QT RT 
 square of tangents=products of intercepts
  2
c c b c a  
 2 = 's
 common SPT RPT    

95. SPT RPT    
SPT TSP  
1 1 1
( ) Hence show thatii
a b c
 
is isoscelesTPS
ST c   = sides in isosceles 
2
PT QT RT 
 square of tangents=products of intercepts
  2
c c b c a  
2 2
c c ac bc ab   
 2 = 's
 common SPT RPT    

96. SPT RPT    
SPT TSP  
1 1 1
( ) Hence show thatii
a b c
 
is isoscelesTPS
ST c   = sides in isosceles 
2
PT QT RT 
 square of tangents=products of intercepts
  2
c c b c a  
2 2
c c ac bc ab   
bc ac ab 
 2 = 's
 common SPT RPT    

97. SPT RPT    
SPT TSP  
1 1 1
( ) Hence show thatii
a b c
 
is isoscelesTPS
ST c   = sides in isosceles 
2
PT QT RT 
 square of tangents=products of intercepts
  2
c c b c a  
2 2
c c ac bc ab   
bc ac ab 
1 1 1
a b c
 
 2 = 's
 common SPT RPT    

98. SPT RPT    
SPT TSP  
1 1 1
( ) Hence show thatii
a b c
 
is isoscelesTPS
ST c   = sides in isosceles 
2
PT QT RT 
 square of tangents=products of intercepts
  2
c c b c a  
2 2
c c ac bc ab   
bc ac ab 
1 1 1
a b c
 
Past HSC Papers
Exercise 10C*
 2 = 's
 common SPT RPT    