x2 t08 01 circle geometry (2012)
TRANSCRIPT
![Page 1: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/1.jpg)
Harder Extension 1 Circle Geometry
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Harder Extension 1
Converse Circle Theorems Circle Geometry
![Page 3: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/3.jpg)
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.
![Page 4: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/4.jpg)
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
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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
![Page 6: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/6.jpg)
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
90 semicircle ain
![Page 7: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/7.jpg)
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
90 semicircle ain
(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.
![Page 8: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/8.jpg)
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
90 semicircle ain
(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
![Page 9: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/9.jpg)
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
90 semicircle ain
(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
![Page 10: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/10.jpg)
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
90 semicircle ain
(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
aresegment samein s
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(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.
![Page 12: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/12.jpg)
(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
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(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.
![Page 14: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/14.jpg)
(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.
![Page 15: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/15.jpg)
(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
![Page 16: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/16.jpg)
(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
![Page 17: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/17.jpg)
(2) The perpendicular bisectors of the sides are concurrent at the
circumcentre which is the centre of the circumcircle, passing
through all three vertices.
![Page 18: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/18.jpg)
(2) The perpendicular bisectors of the sides are concurrent at the
circumcentre which is the centre of the circumcircle, passing
through all three vertices.
![Page 19: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/19.jpg)
(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
![Page 20: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/20.jpg)
(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
![Page 21: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/21.jpg)
(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.
![Page 22: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/22.jpg)
(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.
![Page 23: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/23.jpg)
(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
![Page 24: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/24.jpg)
(4) The altitudes are concurrent at the orthocentre.
![Page 25: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/25.jpg)
(4) The altitudes are concurrent at the orthocentre.
![Page 26: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/26.jpg)
(4) The altitudes are concurrent at the orthocentre.
orthocentre
![Page 27: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/27.jpg)
(4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
![Page 28: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/28.jpg)
(4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
lecircumcirc ifdiameter sinsinsin
C
c
B
b
A
a
![Page 29: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/29.jpg)
(4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
lecircumcirc ifdiameter sinsinsin
C
c
B
b
A
a
Proof: A
B
C
![Page 30: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/30.jpg)
(4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
lecircumcirc ifdiameter sinsinsin
C
c
B
b
A
a
Proof: A
B
C
O
P
![Page 31: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/31.jpg)
(4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
lecircumcirc ifdiameter sinsinsin
C
c
B
b
A
a
Proof: A
B
C
O
P PA segment samein
![Page 32: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/32.jpg)
(4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
lecircumcirc ifdiameter sinsinsin
C
c
B
b
A
a
Proof: A
B
C
O
P PA segment samein
PA sinsin
![Page 33: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/33.jpg)
(4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
lecircumcirc ifdiameter sinsinsin
C
c
B
b
A
a
Proof: A
B
C
O
P PA segment samein
PA sinsin
90PBC
![Page 34: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/34.jpg)
(4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
lecircumcirc ifdiameter sinsinsin
C
c
B
b
A
a
Proof: A
B
C
O
P PA segment samein
PA sinsin
90PBC 90semicirclein
![Page 35: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/35.jpg)
(4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
lecircumcirc ifdiameter sinsinsin
C
c
B
b
A
a
Proof: A
B
C
O
P PA segment samein
PA sinsin
90PBC 90semicirclein
PPC
BCsin
![Page 36: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/36.jpg)
(4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
lecircumcirc ifdiameter sinsinsin
C
c
B
b
A
a
Proof: A
B
C
O
P PA segment samein
PA sinsin
90PBC 90semicirclein
PPC
BCsin
PCP
BC
sin
![Page 37: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/37.jpg)
(4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
lecircumcirc ifdiameter sinsinsin
C
c
B
b
A
a
Proof: A
B
C
O
P PA segment samein
PA sinsin
90PBC 90semicirclein
PPC
BCsin
PCP
BC
sin A
BCPC
sin
![Page 38: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/38.jpg)
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.
![Page 39: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/39.jpg)
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.
![Page 40: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/40.jpg)
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
![Page 41: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/41.jpg)
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
![Page 42: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/42.jpg)
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
ralquadrilate cyclic a is ABCD
![Page 43: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/43.jpg)
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
ralquadrilate cyclic a is ABCD aresegment samein s
![Page 44: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/44.jpg)
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
ralquadrilate cyclic a is ABCD aresegment samein s
90 semicirclein asdiameter is AB
![Page 45: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/45.jpg)
(ii) Show that triangles PCD and APB are similar
![Page 46: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/46.jpg)
(ii) Show that triangles PCD and APB are similar
DPCAPB
![Page 47: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/47.jpg)
(ii) Show that triangles PCD and APB are similar
DPCAPB scommon
![Page 48: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/48.jpg)
(ii) Show that triangles PCD and APB are similar
DPCAPB scommon
PBAPDC
![Page 49: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/49.jpg)
(ii) Show that triangles PCD and APB are similar
DPCAPB scommon
PBAPDC ralquadrilate cyclic exterior
![Page 50: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/50.jpg)
(ii) Show that triangles PCD and APB are similar
DPCAPB scommon
PBAPDC ralquadrilate cyclic exterior
PBAPDC |||
![Page 51: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/51.jpg)
(ii) Show that triangles PCD and APB are similar
DPCAPB scommon
PBAPDC ralquadrilate cyclic exterior
PBAPDC ||| requiangula
![Page 52: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/52.jpg)
(iii) Show that as P varies, the segment CD has constant length.
![Page 53: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/53.jpg)
(iii) Show that as P varies, the segment CD has constant length.
P
A B
P
C D
![Page 54: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/54.jpg)
(iii) Show that as P varies, the segment CD has constant length.
AP
PC
AB
CD
P
A B
P
C D
![Page 55: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/55.jpg)
(iii) Show that as P varies, the segment CD has constant length.
AP
PC
AB
CD s |||in sides of ratio
P
A B
P
C D
![Page 56: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/56.jpg)
(iii) Show that as P varies, the segment CD has constant length.
AP
PC
AB
CD s |||in sides of ratio
PAP
PCPCA cos,In
P
A B
P
C D
![Page 57: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/57.jpg)
(iii) Show that as P varies, the segment CD has constant length.
AP
PC
AB
CD s |||in sides of ratio
PAP
PCPCA cos,In
PAB
CDcos
P
A B
P
C D
![Page 58: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/58.jpg)
(iii) Show that as P varies, the segment CD has constant length.
AP
PC
AB
CD s |||in sides of ratio
PAP
PCPCA cos,In
PAB
CDcos
PABCD cos
P
A B
P
C D
![Page 59: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/59.jpg)
(iii) Show that as P varies, the segment CD has constant length.
AP
PC
AB
CD s |||in sides of ratio
PAP
PCPCA cos,In
PAB
CDcos
PABCD cos
constant is Now, P
P
A B
P
C D
![Page 60: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/60.jpg)
(iii) Show that as P varies, the segment CD has constant length.
AP
PC
AB
CD s |||in sides of ratio
PAP
PCPCA cos,In
PAB
CDcos
PABCD cos
constant is Now, P
P
A B
P
C D
aresegment samein s
![Page 61: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/61.jpg)
(iii) Show that as P varies, the segment CD has constant length.
AP
PC
AB
CD s |||in sides of ratio
PAP
PCPCA cos,In
PAB
CDcos
PABCD cos
constant is Now, P
P
A B
P
C D
fixed is and AB
aresegment samein s
![Page 62: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/62.jpg)
(iii) Show that as P varies, the segment CD has constant length.
AP
PC
AB
CD s |||in sides of ratio
PAP
PCPCA cos,In
PAB
CDcos
PABCD cos
constant is Now, P
P
A B
P
C D
fixed is and AB
aresegment samein s
given
![Page 63: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/63.jpg)
(iii) Show that as P varies, the segment CD has constant length.
AP
PC
AB
CD s |||in sides of ratio
PAP
PCPCA cos,In
PAB
CDcos
PABCD cos
constant is Now, P
P
A B
P
C D
fixed is and AB
aresegment samein s
givenconstant is CD
![Page 64: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/64.jpg)
(iv) Find the locus of the midpoint of CD.
![Page 65: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/65.jpg)
(iv) Find the locus of the midpoint of CD.
ABCD is a cyclic quadrilateral with AB diameter.
A B
C D
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(iv) Find the locus of the midpoint of CD.
ABCD is a cyclic quadrilateral with AB diameter.
A B
C D Let M be the midpoint of CD M
![Page 67: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/67.jpg)
(iv) Find the locus of the midpoint of CD.
ABCD is a cyclic quadrilateral with AB diameter.
A B
C D Let M be the midpoint of CD
O is the midpoint of AB
M
O
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(iv) Find the locus of the midpoint of CD.
ABCD is a cyclic quadrilateral with AB diameter.
A B
C D Let M be the midpoint of CD
O is the midpoint of AB
M
O
OM is constant
![Page 69: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/69.jpg)
(iv) Find the locus of the midpoint of CD.
ABCD is a cyclic quadrilateral with AB diameter.
A B
C D Let M be the midpoint of CD
O is the midpoint of AB
M
O
OM is constant
centre thefromt equidistan are chords
![Page 70: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/70.jpg)
(iv) Find the locus of the midpoint of CD.
ABCD is a cyclic quadrilateral with AB diameter.
A B
C D Let M be the midpoint of CD
O is the midpoint of AB
M
O
OM is constant
centre thefromt equidistan are chords
OM from distance fixed a is
![Page 71: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/71.jpg)
(iv) Find the locus of the midpoint of CD.
ABCD is a cyclic quadrilateral with AB diameter.
A B
C D Let M be the midpoint of CD
O is the midpoint of AB
M
O
OM is constant
centre thefromt equidistan are chords
OM from distance fixed a is 222 MCOCOM
![Page 72: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/72.jpg)
(iv) Find the locus of the midpoint of CD.
ABCD is a cyclic quadrilateral with AB diameter.
A B
C D Let M be the midpoint of CD
O is the midpoint of AB
M
O
OM is constant
centre thefromt equidistan are chords
OM from distance fixed a is 222 MCOCOM
PAB
PABAB
PABAB
22
222
22
sin4
1
cos4
1
4
1
cos2
1
2
1
![Page 73: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/73.jpg)
(iv) Find the locus of the midpoint of CD.
ABCD is a cyclic quadrilateral with AB diameter.
A B
C D Let M be the midpoint of CD
O is the midpoint of AB
M
O
OM is constant
centre thefromt equidistan are chords
OM from distance fixed a is 222 MCOCOM
PAB
PABAB
PABAB
22
222
22
sin4
1
cos4
1
4
1
cos2
1
2
1
PABOM sin2
1
![Page 74: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/74.jpg)
(iv) Find the locus of the midpoint of CD.
ABCD is a cyclic quadrilateral with AB diameter.
A B
C D Let M be the midpoint of CD
O is the midpoint of AB
M
O
OM is constant
centre thefromt equidistan are chords
OM from distance fixed a is 222 MCOCOM
PAB
PABAB
PABAB
22
222
22
sin4
1
cos4
1
4
1
cos2
1
2
1
PABOM sin2
1
PAB
O
sin2
1radius and
centre circle, is locus
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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
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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 that i TSP TPS
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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 that i TSP TPS
RQP RPT
![Page 78: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/78.jpg)
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 that i TSP TPS
RQP RPT alternate segment theorem
![Page 79: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/79.jpg)
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 that i TSP TPS
RQP RPT alternate segment theorem
TSP RQP SPQ
![Page 80: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/80.jpg)
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 that i TSP TPS
RQP RPT alternate segment theorem
TSP RQP SPQ exterior , SPQ
![Page 81: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/81.jpg)
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 that i TSP TPS
RQP RPT alternate segment theorem
TSP RQP SPQ exterior , SPQ
TSP RPT
![Page 82: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/82.jpg)
SPT RPT
![Page 83: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/83.jpg)
SPT RPT common
![Page 84: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/84.jpg)
SPT RPT common
SPT TSP
![Page 85: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/85.jpg)
SPT RPT common
SPT TSP
1 1 1( ) Hence show that ii
a b c
![Page 86: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/86.jpg)
SPT RPT common
SPT TSP
1 1 1( ) Hence show that ii
a b c
![Page 87: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/87.jpg)
SPT RPT common
SPT TSP
1 1 1( ) Hence show that ii
a b c
is isoscelesTPS
![Page 88: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/88.jpg)
SPT RPT
common
2 = 's
SPT TSP
1 1 1( ) Hence show that ii
a b c
is isoscelesTPS
SPT RPT
![Page 89: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/89.jpg)
SPT RPT
SPT TSP
1 1 1( ) Hence show that ii
a b c
is isoscelesTPS 2 = 's
common SPT RPT
![Page 90: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/90.jpg)
SPT RPT
SPT TSP
1 1 1( ) Hence show that ii
a b c
is isoscelesTPS
ST c
2 = 's
common SPT RPT
![Page 91: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/91.jpg)
SPT RPT
SPT TSP
1 1 1( ) Hence show that ii
a b c
is isoscelesTPS
ST c = sides in isosceles
2 = 's
common SPT RPT
![Page 92: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/92.jpg)
SPT RPT
SPT TSP
1 1 1( ) Hence show that ii
a b c
is isoscelesTPS
ST c = sides in isosceles 2PT QT RT
2 = 's
common SPT RPT
![Page 93: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/93.jpg)
SPT RPT
SPT TSP
1 1 1( ) Hence show that ii
a b c
is isoscelesTPS
ST c = sides in isosceles 2PT QT RT
square of tangents=products of intercepts
2 = 's
common SPT RPT
![Page 94: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/94.jpg)
SPT RPT
SPT TSP
1 1 1( ) Hence show that ii
a b c
is isoscelesTPS
ST c = sides in isosceles 2PT QT RT
square of tangents=products of intercepts
2c c b c a
2 = 's
common SPT RPT
![Page 95: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/95.jpg)
SPT RPT
SPT TSP
1 1 1( ) Hence show that ii
a b c
is isoscelesTPS
ST c = sides in isosceles 2PT QT RT
square of tangents=products of intercepts
2c c b c a
2 2c c ac bc ab
2 = 's
common SPT RPT
![Page 96: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/96.jpg)
SPT RPT
SPT TSP
1 1 1( ) Hence show that ii
a b c
is isoscelesTPS
ST c = sides in isosceles 2PT QT RT
square of tangents=products of intercepts
2c c b c a
2 2c c ac bc ab
bc ac ab
2 = 's
common SPT RPT
![Page 97: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/97.jpg)
SPT RPT
SPT TSP
1 1 1( ) Hence show that ii
a b c
is isoscelesTPS
ST c = sides in isosceles 2PT QT RT
square of tangents=products of intercepts
2c c b c a
2 2c c ac bc ab
bc ac ab
1 1 1
a b c
2 = 's
common SPT RPT
![Page 98: X2 t08 01 circle geometry (2012)](https://reader036.vdocuments.net/reader036/viewer/2022081404/55911a141a28ab9b758b4661/html5/thumbnails/98.jpg)
SPT RPT
SPT TSP
1 1 1( ) Hence show that ii
a b c
is isoscelesTPS
ST c = sides in isosceles 2PT QT RT
square of tangents=products of intercepts
2c c b c a
2 2c c ac bc ab
bc ac ab
1 1 1
a b c
Past HSC Papers
Exercise 10C*
2 = 's
common SPT RPT