X2 T08 01 circle geometry
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Harder Extension 1 Circle Geometry
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Transcript of X2 T08 01 circle geometry
Harder Extension 1Circle Geometry
Harder Extension 1Converse Circle Theorems
Circle Geometry
Harder Extension 1Converse Circle Theorems
Circle Geometry(1) The circle whose diameter is the hypotenuse of a right angled
triangle passes through the third vertex.
Harder Extension 1Converse 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 1Converse Circle Theorems
Circle Geometry(1) The circle whose diameter is the hypotenuse of a right angled
triangle passes through the third vertex.
A B
CABC are concyclic with AB diameter
Harder Extension 1Converse Circle Theorems
Circle Geometry(1) The circle whose diameter is the hypotenuse of a right angled
triangle passes through the third vertex.
A B
CABC are concyclic with AB diameter
90 semicircle ain
Harder Extension 1Converse Circle Theorems
Circle Geometry(1) The circle whose diameter is the hypotenuse of a right angled
triangle passes through the third vertex.
A B
CABC 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 1Converse Circle Theorems
Circle Geometry(1) The circle whose diameter is the hypotenuse of a right angled
triangle passes through the third vertex.
A B
CABC 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 1Converse Circle Theorems
Circle Geometry(1) The circle whose diameter is the hypotenuse of a right angled
triangle passes through the third vertex.
A B
CABC 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 1Converse Circle Theorems
Circle Geometry(1) The circle whose diameter is the hypotenuse of a right angled
triangle passes through the third vertex.
A B
CABC 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.
orthocentreInteraction Between Geometry & Trigonometry
(4) The altitudes are concurrent at the orthocentre.
orthocentreInteraction Between Geometry & Trigonometry
lecircumcirc ifdiameter sinsinsin
C
cB
bA
a
(4) The altitudes are concurrent at the orthocentre.
orthocentreInteraction Between Geometry & Trigonometry
lecircumcirc ifdiameter sinsinsin
C
cB
bA
a
Proof: A
B
C
(4) The altitudes are concurrent at the orthocentre.
orthocentreInteraction Between Geometry & Trigonometry
lecircumcirc ifdiameter sinsinsin
C
cB
bA
a
Proof: A
B
C
O
P
(4) The altitudes are concurrent at the orthocentre.
orthocentreInteraction Between Geometry & Trigonometry
lecircumcirc ifdiameter sinsinsin
C
cB
bA
a
Proof: A
B
C
O
PPA segment samein
(4) The altitudes are concurrent at the orthocentre.
orthocentreInteraction Between Geometry & Trigonometry
lecircumcirc ifdiameter sinsinsin
C
cB
bA
a
Proof: A
B
C
O
PPA segment samein PA sinsin
(4) The altitudes are concurrent at the orthocentre.
orthocentreInteraction Between Geometry & Trigonometry
lecircumcirc ifdiameter sinsinsin
C
cB
bA
a
Proof: A
B
C
O
PPA segment samein PA sinsin 90PBC
(4) The altitudes are concurrent at the orthocentre.
orthocentreInteraction Between Geometry & Trigonometry
lecircumcirc ifdiameter sinsinsin
C
cB
bA
a
Proof: A
B
C
O
PPA segment samein PA sinsin 90PBC 90semicirclein
(4) The altitudes are concurrent at the orthocentre.
orthocentreInteraction Between Geometry & Trigonometry
lecircumcirc ifdiameter sinsinsin
C
cB
bA
a
Proof: A
B
C
O
PPA segment samein PA sinsin 90PBC 90semicirclein
PPCBC sin
(4) The altitudes are concurrent at the orthocentre.
orthocentreInteraction Between Geometry & Trigonometry
lecircumcirc ifdiameter sinsinsin
C
cB
bA
a
Proof: A
B
C
O
PPA segment samein PA sinsin 90PBC 90semicirclein
PPCBC sin
PCP
BC
sin
(4) The altitudes are concurrent at the orthocentre.
orthocentreInteraction Between Geometry & Trigonometry
lecircumcirc ifdiameter sinsinsin
C
cB
bA
a
Proof: A
B
C
O
PPA segment samein PA sinsin 90PBC 90semicirclein
PPCBC sin
PCP
BC
sin ABCPC
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
ralquadrilatecyclic 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
ralquadrilatecyclic 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
ralquadrilatecyclic a is ABCD aresegment samein s90semicirclein 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 ralquadrilatecyclicexterior
(ii) Show that triangles PCD and APB are similar
DPCAPB scommon
PBAPDC ralquadrilatecyclicexterior
PBAPDC |||
(ii) Show that triangles PCD and APB are similar
DPCAPB scommon
PBAPDC ralquadrilatecyclicexterior
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.
APPC
ABCD
P
A B
P
C D
(iii) Show that as P varies, the segment CD has constant length.
APPC
ABCD
s|||in sidesofratio
P
A B
P
C D
(iii) Show that as P varies, the segment CD has constant length.
APPC
ABCD
s|||in sidesofratio
PAPPCPCA cos,In
P
A B
P
C D
(iii) Show that as P varies, the segment CD has constant length.
APPC
ABCD
s|||in sidesofratio
PAPPCPCA cos,In
PABCD cos
P
A B
P
C D
(iii) Show that as P varies, the segment CD has constant length.
APPC
ABCD
s|||in sidesofratio
PAPPCPCA cos,In
PABCD cos
PABCD cos
P
A B
P
C D
(iii) Show that as P varies, the segment CD has constant length.
APPC
ABCD
s|||in sidesofratio
PAPPCPCA cos,In
PABCD cos
PABCD cosconstant is Now, P
P
A B
P
C D
(iii) Show that as P varies, the segment CD has constant length.
APPC
ABCD
s|||in sidesofratio
PAPPCPCA cos,In
PABCD cos
PABCD cosconstant is Now, P
P
A B
P
C D
aresegment samein s
(iii) Show that as P varies, the segment CD has constant length.
APPC
ABCD
s|||in sidesofratio
PAPPCPCA cos,In
PABCD cos
PABCD cosconstant 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.
APPC
ABCD
s|||in sidesofratio
PAPPCPCA cos,In
PABCD cos
PABCD cosconstant 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.
APPC
ABCD
s|||in sidesofratio
PAPPCPCA cos,In
PABCD cos
PABCD cosconstant is Now, P
P
A B
P
C D
fixed is and AB aresegment samein s given
constant 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
CD
(iv) Find the locus of the midpoint of CD.ABCD is a cyclic quadrilateral with AB diameter.
A B
CD Let M be the midpoint of CDM
(iv) Find the locus of the midpoint of CD.ABCD is a cyclic quadrilateral with AB diameter.
A B
CD Let M be the midpoint of CDO 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
CD Let M be the midpoint of CDO is the midpoint of AB
M
OOM is constant
(iv) Find the locus of the midpoint of CD.ABCD is a cyclic quadrilateral with AB diameter.
A B
CD Let M be the midpoint of CDO is the midpoint of AB
M
OOM is constant centrethefromt equidistan arechords
(iv) Find the locus of the midpoint of CD.ABCD is a cyclic quadrilateral with AB diameter.
A B
CD Let M be the midpoint of CDO is the midpoint of AB
M
OOM is constant centrethefromt equidistan arechords
OM fromdistance fixedais
(iv) Find the locus of the midpoint of CD.ABCD is a cyclic quadrilateral with AB diameter.
A B
CD Let M be the midpoint of CDO is the midpoint of AB
M
OOM is constant centrethefromt equidistan arechords
OM fromdistance fixedais 222 MCOCOM
(iv) Find the locus of the midpoint of CD.ABCD is a cyclic quadrilateral with AB diameter.
A B
CD Let M be the midpoint of CDO is the midpoint of AB
M
OOM is constant centrethefromt equidistan arechords
OM fromdistance fixedais 222 MCOCOM
PAB
PABAB
PABAB
22
222
22
sin41
cos41
41
cos21
21
(iv) Find the locus of the midpoint of CD.ABCD is a cyclic quadrilateral with AB diameter.
A B
CD Let M be the midpoint of CDO is the midpoint of AB
M
OOM is constant centrethefromt equidistan arechords
OM fromdistance fixedais 222 MCOCOM
PAB
PABAB
PABAB
22
222
22
sin41
cos41
41
cos21
21
PABOM sin21
(iv) Find the locus of the midpoint of CD.ABCD is a cyclic quadrilateral with AB diameter.
A B
CD Let M be the midpoint of CDO is the midpoint of AB
M
OOM is constant centrethefromt equidistan arechords
OM fromdistance fixedais 222 MCOCOM
PAB
PABAB
PABAB
22
222
22
sin41
cos41
41
cos21
21
PABOM sin21
PAB
O
sin21radius and
centrecircle,islocus
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.
PQRQPS 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.
PQRQPS 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.
PQRQPS 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.
PQRQPS 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.
PQRQPS RPS
( ) Show that i TSP TPS RQP RPT alternate segment theoremTSP 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.
PQRQPS RPS
( ) Show that i TSP TPS RQP RPT alternate segment theoremTSP 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.
PQRQPS RPS
( ) Show that i TSP TPS RQP RPT alternate segment theoremTSP 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 iia b c
SPT RPT common SPT TSP
1 1 1( ) Hence show that iia b c
SPT RPT common SPT TSP
1 1 1( ) Hence show that iia b c
is isoscelesTPS
SPT RPT
common
2 = 's
SPT TSP
1 1 1( ) Hence show that iia b c
is isoscelesTPS
SPT RPT
SPT RPT
SPT TSP
1 1 1( ) Hence show that iia b c
is isoscelesTPS 2 = 's
common SPT RPT
SPT RPT
SPT TSP
1 1 1( ) Hence show that iia b c
is isoscelesTPSST c
2 = 's
common SPT RPT
SPT RPT
SPT TSP
1 1 1( ) Hence show that iia b c
is isoscelesTPSST c = sides in isosceles
2 = 's
common SPT RPT
SPT RPT
SPT TSP
1 1 1( ) Hence show that iia b c
is isoscelesTPSST c = sides in isosceles 2PT QT RT
2 = 's
common SPT RPT
SPT RPT
SPT TSP
1 1 1( ) Hence show that iia b c
is isoscelesTPSST 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 iia b c
is isoscelesTPSST 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 iia b c
is isoscelesTPSST 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 iia b c
is isoscelesTPSST 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 iia b c
is isoscelesTPSST 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 1a b c
2 = 's
common SPT RPT
SPT RPT
SPT TSP
1 1 1( ) Hence show that iia b c
is isoscelesTPSST 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 1a b c
Past HSC Papers
Exercise 10C*
2 = 's
common SPT RPT
