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- 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 A P in same segment P 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 A P in same segment P 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 A P in same segment P 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 A P in same segment P 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 A P in same segment P 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 A P in same segment P 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 A P in same segment P 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. M D 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. M D 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. M D 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. M D C Let M be the midpoint of CD O is the midpoint of AB OM is constant A O B chords are equidistant from the centre
- 70. (iv) Find the locus of the midpoint of CD. ABCD is a cyclic quadrilateral with AB diameter. M D C Let M be the midpoint of CD O is the midpoint of AB OM is constant A O B 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. M D C Let M be the midpoint of CD O is the midpoint of AB OM is constant A O B 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. M D C Let M be the midpoint of CD O is the midpoint of AB OM is constant A O B chords are equidistant from the centre M is a fixed distance from O OM 2 OC 2 MC 2 2 2 1 1 AB AB cos P 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. M D C Let M be the midpoint of CD O is the midpoint of AB OM is constant A O B chords are equidistant from the centre M is a fixed distance from O OM 2 OC 2 MC 2 2 2 1 1 AB AB cos P 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. M D C Let M be the midpoint of CD O is the midpoint of AB OM is constant A O B 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 1 1 AB AB cos P 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*

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