x2 t08 01 circle geometry (2012)
TRANSCRIPT
Harder Extension 1 Circle Geometry
Harder Extension 1
Converse Circle Theorems Circle Geometry
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.
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
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
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
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.
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
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
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
(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.
(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
(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.
(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.
(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
(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
(2) The perpendicular bisectors of the sides are concurrent at the
circumcentre which is the centre of the circumcircle, passing
through all three vertices.
(2) The perpendicular bisectors of the sides are concurrent at the
circumcentre which is the centre of the circumcircle, passing
through all three vertices.
(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
(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
(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.
(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.
(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
(4) The altitudes are concurrent at the orthocentre.
(4) The altitudes are concurrent at the orthocentre.
(4) The altitudes are concurrent at the orthocentre.
orthocentre
(4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
(4) The altitudes are concurrent at the orthocentre.
orthocentre
Interaction Between Geometry & Trigonometry
lecircumcirc ifdiameter sinsinsin
C
c
B
b
A
a
(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
(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
(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
(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
(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
(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
(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
(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
(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
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.
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.
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
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
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
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
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
(ii) Show that triangles PCD and APB are similar
(ii) Show that triangles PCD and APB are similar
DPCAPB
(ii) Show that triangles PCD and APB are similar
DPCAPB scommon
(ii) Show that triangles PCD and APB are similar
DPCAPB scommon
PBAPDC
(ii) Show that triangles PCD and APB are similar
DPCAPB scommon
PBAPDC ralquadrilate cyclic exterior
(ii) Show that triangles PCD and APB are similar
DPCAPB scommon
PBAPDC ralquadrilate cyclic exterior
PBAPDC |||
(ii) Show that triangles PCD and APB are similar
DPCAPB scommon
PBAPDC ralquadrilate cyclic exterior
PBAPDC ||| requiangula
(iii) Show that as P varies, the segment CD has constant length.
(iii) Show that as P varies, the segment CD has constant length.
P
A B
P
C D
(iii) Show that as P varies, the segment CD has constant length.
AP
PC
AB
CD
P
A B
P
C D
(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
(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
(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
(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
(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
(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
(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
(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
(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
(iv) Find the locus of the midpoint of CD.
(iv) Find the locus of the midpoint of CD.
ABCD is a cyclic quadrilateral with AB diameter.
A B
C D
(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
(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
(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
(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
(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
(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
(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
(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
(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
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
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
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
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
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
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
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
SPT RPT
SPT RPT common
SPT RPT common
SPT TSP
SPT RPT common
SPT TSP
1 1 1( ) Hence show that ii
a b c
SPT RPT common
SPT TSP
1 1 1( ) Hence show that ii
a b c
SPT RPT common
SPT TSP
1 1 1( ) Hence show that ii
a b c
is isoscelesTPS
SPT RPT
common
2 = 's
SPT TSP
1 1 1( ) Hence show that ii
a b c
is isoscelesTPS
SPT RPT
SPT RPT
SPT TSP
1 1 1( ) Hence show that ii
a b c
is isoscelesTPS 2 = 's
common SPT RPT
SPT RPT
SPT TSP
1 1 1( ) Hence show that ii
a b c
is isoscelesTPS
ST c
2 = 's
common SPT RPT
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
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
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
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
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
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
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
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