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
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
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
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
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*