1. Harder Extension 1
Circle Geometry

2. Harder Extension 1
Circle Geometry
Converse Circle Theorems

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

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

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

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

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

9. Harder Extension 1
Circle Geometry
Converse Circle Theorems
(1) The circle whose diameter is the hypotenuse of a right angled
triangle passes through the third vertex.
C
ABC are concyclic with AB diameter
 in a semicircle  90 

A B
(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.
P Q
  ABQP is a cyclic quadrilateral
A B

10. Harder Extension 1
Circle Geometry
Converse Circle Theorems
(1) The circle whose diameter is the hypotenuse of a right angled
triangle passes through the third vertex.
C
ABC are concyclic with AB diameter
 in a semicircle  90 

A B
(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.
P Q
  ABQP is a cyclic quadrilateral
s in same segment are  
A B

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
a b c
   diameter if circumcircle
sin A sin B sin C

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

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

31. (4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
a b c
   diameter if circumcircle
sin A sin B sin C
Proof: A P A  P  in same segment  
O B
C

32. (4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
a b c
   diameter if circumcircle
sin A sin B sin C
Proof: A P A  P  in same segment  
 sin A  sin P
O B
C

33. (4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
a b c
   diameter if circumcircle
sin A sin B sin C
Proof: A P A  P  in same segment  
 sin A  sin P
PBC  90
O B
C

34. (4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
a b c
   diameter if circumcircle
sin A sin B sin C
Proof: A P A  P  in same segment  
 sin A  sin P
PBC  90  in semicircle  90 

O B
C

35. (4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
a b c
   diameter if circumcircle
sin A sin B sin C
Proof: A P A  P  in same segment  
 sin A  sin P
PBC  90  in semicircle  90 

BC
O B  sin P
PC
C

36. (4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
a b c
   diameter if circumcircle
sin A sin B sin C
Proof: A P A  P  in same segment  
 sin A  sin P
PBC  90  in semicircle  90 

BC
O B  sin P
PC
BC
 PC
C sin P

37. (4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
a b c
   diameter if circumcircle
sin A sin B sin C
Proof: A P A  P  in same segment  
 sin A  sin P
PBC  90  in semicircle  90 

BC
O B  sin P
PC
BC BC
 PC  PC 
sin P sin A
C

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.
BDA  ACB  90

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.
BDA  ACB  90 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.
BDA  ACB  90 given 
 ABCD is a cyclic quadrilateral

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.
BDA  ACB  90 given 
 ABCD is a cyclic quadrilateral s in same segment are  

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.
BDA  ACB  90 given 
 ABCD is a cyclic quadrilateral s in same segment are  
AB is diameter as  in semicircle  90

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

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

47. (ii) Show that triangles PCD and APB are similar
APB  DPC common s 

48. (ii) Show that triangles PCD and APB are similar
APB  DPC common s 
PDC  PBA

49. (ii) Show that triangles PCD and APB are similar
APB  DPC common s 
PDC  PBA exterior cyclic quadrilateral

50. (ii) Show that triangles PCD and APB are similar
APB  DPC common s 
PDC  PBA exterior cyclic quadrilateral
 PDC ||| PBA

51. (ii) Show that triangles PCD and APB are similar
APB  DPC common s 
PDC  PBA exterior cyclic quadrilateral
 PDC ||| PBA equiangular 

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

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

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

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

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

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

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

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

60. P P
A B C D
(iii) Show that as P varies, the segment CD has constant length.
CD PC
 ratio of sides in ||| s 
AB AP
PC
In PCA,  cos P
AP
CD
  cos P
AB
CD  AB cos P
Now, P is constant s in same segment are  

61. P P
A B C D
(iii) Show that as P varies, the segment CD has constant length.
CD PC
 ratio of sides in ||| s 
AB AP
PC
In PCA,  cos P
AP
CD
  cos P
AB
CD  AB cos P
Now, P is constant s in same segment are  
and AB is fixed

62. P P
A B C D
(iii) Show that as P varies, the segment CD has constant length.
CD PC
 ratio of sides in ||| s 
AB AP
PC
In PCA,  cos P
AP
CD
  cos P
AB
CD  AB cos P
Now, P is constant s in same segment are  
and AB is fixed given 

63. P P
A B C D
(iii) Show that as P varies, the segment CD has constant length.
CD PC
 ratio of sides in ||| s 
AB AP
PC
In PCA,  cos P
AP
CD
  cos P
AB
CD  AB cos P
Now, P is constant s in same segment are  
and AB is fixed given 
 CD is constant

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.
D C
A B

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

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

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

69. (iv) Find the locus of the midpoint of CD.
ABCD is a cyclic quadrilateral with AB diameter.
D M C Let M be the midpoint of CD
O is the midpoint of AB
OM is constant
A B
O  chords are equidistant from the centre

70. (iv) Find the locus of the midpoint of CD.
ABCD is a cyclic quadrilateral with AB diameter.
D M C Let M be the midpoint of CD
O is the midpoint of AB
OM is constant
A B
O  chords are equidistant from the centre
 M is a fixed distance from O

71. (iv) Find the locus of the midpoint of CD.
ABCD is a cyclic quadrilateral with AB diameter.
D M C Let M be the midpoint of CD
O is the midpoint of AB
OM is constant
A B
O  chords are equidistant from the centre
 M is a fixed distance from O
OM 2  OC 2  MC 2

72. (iv) Find the locus of the midpoint of CD.
ABCD is a cyclic quadrilateral with AB diameter.
D M C Let M be the midpoint of CD
O is the midpoint of AB
OM is constant
A B
O  chords are equidistant from the centre
 M is a fixed distance from O
OM 2  OC 2  MC 2
2 2
  AB    AB cos P 
1 1
   
2  2 
1 1
 AB  AB 2 cos 2 P
2
4 4
1
 AB 2 sin 2 P
4

73. (iv) Find the locus of the midpoint of CD.
ABCD is a cyclic quadrilateral with AB diameter.
D M C Let M be the midpoint of CD
O is the midpoint of AB
OM is constant
A B
O  chords are equidistant from the centre
 M is a fixed distance from O
OM 2  OC 2  MC 2
2 2
  AB    AB cos P 
1 1
   
2  2 
1 1
 AB  AB 2 cos 2 P
2
4 4
1
 AB 2 sin 2 P
4
1
OM  AB sin P
2

74. (iv) Find the locus of the midpoint of CD.
ABCD is a cyclic quadrilateral with AB diameter.
D M C Let M be the midpoint of CD
O is the midpoint of AB
OM is constant
A B
O  chords are equidistant from the centre
 M is a fixed distance from O  locus is circle, centre O
OM 2  OC 2  MC 2 1
2 2 and radius  AB sin P
  AB    AB cos P 
1 1 2
   
2  2 
1 1
 AB  AB 2 cos 2 P
2
4 4
1
 AB 2 sin 2 P
4
1
OM  AB sin P
2

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 PQR meets QR at S
so that QPS  RPS   . The intervals RS, SQ and PT have lengths
a, b and c respectively.

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 PQR meets QR at S
so that QPS  RPS   . The intervals RS, SQ and PT have lengths
a, b and c respectively.
(i ) Show that 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 PQR meets QR at S
so that QPS  RPS   . The intervals RS, SQ and PT have lengths
a, b and c respectively.
(i ) Show that 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 PQR meets QR at S
so that QPS  RPS   . The intervals RS, SQ and PT have lengths
a, b and c respectively.
(i ) Show that 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 PQR meets QR at S
so that QPS  RPS   . The intervals RS, SQ and PT have lengths
a, b and c respectively.
(i ) Show that 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 PQR meets QR at S
so that QPS  RPS   . The intervals RS, SQ and PT have lengths
a, b and c respectively.
(i ) Show that 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 PQR meets QR at S
so that QPS  RPS   . The intervals RS, SQ and PT have lengths
a, b and c respectively.
(i ) Show that 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
(ii ) Hence show that  
a b c

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

87. SPT  RPT    common  
SPT  TSP
1 1 1
(ii ) Hence show that  
a b c
TPS is isosceles

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

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

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

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

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

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

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

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

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

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

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