A theorem is a general proposition that is not self-evident but is proved by reasoning (these proofs need not be learned for examination purposes).
Theorem 1 The line drawn from the centre of a circle, perpendicular to a chord, bisects the chord.
Proof:
Consider a circle, with centre OO. Draw a chord ABAB and draw a perpendicular line from the centre of the circle to intersect the chord at point PP.
The aim is to prove that APAP = BPBP
- ▵OAP▵OAP and ▵OBP▵OBP are right-angled triangles.
- OA=OBOA=OB as both of these are radii and OPOP is common to both triangles.
Apply the Theorem of Pythagoras to each triangle, to get:
O
A
2
=
O
P
2
+
A
P
2
O
B
2
=
O
P
2
+
B
P
2
O
A
2
=
O
P
2
+
A
P
2
O
B
2
=
O
P
2
+
B
P
2
(1)However, OA=OBOA=OB. So,
O
P
2
+
A
P
2
=
O
P
2
+
B
P
2
∴
A
P
2
=
B
P
2
and AP
=
B
P
O
P
2
+
A
P
2
=
O
P
2
+
B
P
2
∴
A
P
2
=
B
P
2
and AP
=
B
P
(2)This means that OPOP bisects ABAB.
Theorem 2 The line drawn from the centre of a circle, that bisects a chord, is perpendicular to the chord.
Proof:
Consider a circle, with centre OO. Draw a chord ABAB and draw a line from the centre of the circle to bisect the chord at point PP.
The aim is to prove that OP⊥ABOP⊥AB
In ▵OAP▵OAP and ▵OBP▵OBP,
- AP=PBAP=PB (given)
- OA=OBOA=OB (radii)
- OPOP is common to both triangles.
∴▵OAP≡▵OBP∴▵OAP≡▵OBP (SSS).
O
P
A
^
=
O
P
B
^
O
P
A
^
+
O
P
B
^
=
180
∘
(
APB
is a str. line
)
∴
O
P
A
^
=
O
P
B
^
=
90
∘
∴
O
P
⊥
A
B
O
P
A
^
=
O
P
B
^
O
P
A
^
+
O
P
B
^
=
180
∘
(
APB
is a str. line
)
∴
O
P
A
^
=
O
P
B
^
=
90
∘
∴
O
P
⊥
A
B
(3)Theorem 3 The perpendicular bisector of a chord passes through the centre of the circle.
Proof:
Consider a circle. Draw a chord ABAB. Draw a line PQPQ perpendicular to ABAB such that PQPQ bisects ABAB at point PP. Draw lines AQAQ and BQBQ.
The aim is to prove that QQ is the centre of the circle, by showing that AQ=BQAQ=BQ.
In ▵OAP▵OAP and ▵OBP▵OBP,
- AP=PBAP=PB (given)
- ∠QPA=∠QPB∠QPA=∠QPB (QP⊥ABQP⊥AB)
- QPQP is common to both triangles.
∴▵QAP≡▵QBP∴▵QAP≡▵QBP (SAS).
From this, QA=QBQA=QB. Since the centre of a circle is the only point inside a circle that has points on the circumference at an equal distance from it, QQ must be the centre of the circle.
Theorem 4 The angle subtended by an arc at the centre of a circle is double the size of the angle subtended by the same arc at the circumference of the circle.
Proof:
Consider a circle, with centre OO and with AA and BB on the circumference. Draw a chord ABAB. Draw radii OAOA and OBOB. Select any point PP on the circumference of the circle. Draw lines PAPA and PBPB. Draw POPO and extend to RR.
The aim is to prove that AOB^=2·APB^AOB^=2·APB^.AOR^=PAO^+APO^AOR^=PAO^+APO^ (exterior angle = sum of interior opp. angles)
But, PAO^=APO^PAO^=APO^ (▵AOP▵AOP is an isosceles ▵▵)
∴∴AOR^=2APO^AOR^=2APO^
Similarly, BOR^=2BPO^BOR^=2BPO^.
So,
A
O
B
^
=
A
O
R
^
+
B
O
R
^
=
2
A
P
O
^
+
2
B
P
O
^
=
2
(
A
P
O
^
+
B
P
O
^
)
=
2
(
A
P
B
^
)
A
O
B
^
=
A
O
R
^
+
B
O
R
^
=
2
A
P
O
^
+
2
B
P
O
^
=
2
(
A
P
O
^
+
B
P
O
^
)
=
2
(
A
P
B
^
)
(4)- Find the angles (aa to ff) indicated in each diagram:
Theorem 5 The angles subtended by a chord at the circumference of a circle on the same side of the chord are equal.
Proof:
Consider a circle, with centre OO. Draw a chord ABAB. Select any points PP and QQ on the circumference of the circle, such that both PP and QQ are on the same side of the chord. Draw lines PAPA, PBPB, QAQA and QBQB.
The aim is to prove that AQB^=APB^AQB^=APB^.
A
O
B
^
=
2
A
Q
B
^
∠
at centre
=
twice
∠
at circumference
and
AOB
^
=
2
A
P
B
^
∠
at centre
=
twice
∠
at circumference
∴
2
A
Q
B
^
=
2
A
P
B
^
∴
A
Q
B
^
=
A
P
B
^
A
O
B
^
=
2
A
Q
B
^
∠
at centre
=
twice
∠
at circumference
and
AOB
^
=
2
A
P
B
^
∠
at centre
=
twice
∠
at circumference
∴
2
A
Q
B
^
=
2
A
P
B
^
∴
A
Q
B
^
=
A
P
B
^
(5)Theorem 6 (Converse of Theorem (Reference)) If a line segment subtends equal angles at two other points on the same side of the line, then these four points lie on a circle.
Proof:
Consider a line segment ABAB, that subtends equal angles at points PP and QQ on the same side of ABAB.
The aim is to prove that points AA, BB, PP and QQ lie on the circumference of a circle.
By contradiction. Assume that point PP does not lie on a circle drawn through points AA, BB and QQ. Let the circle cut APAP (or APAP extended) at point RR.
A
Q
B
^
=
A
R
B
^
∠
's on same side of chord
but
AQB
^
=
A
P
B
^
(
given
)
∴
A
R
B
^
=
A
P
B
^
but this cannot be true since
ARB
^
=
A
P
B
^
+
R
B
P
^
(
ext.
∠
of
▵
)
A
Q
B
^
=
A
R
B
^
∠
's on same side of chord
but
AQB
^
=
A
P
B
^
(
given
)
∴
A
R
B
^
=
A
P
B
^
but this cannot be true since
ARB
^
=
A
P
B
^
+
R
B
P
^
(
ext.
∠
of
▵
)
(6)∴∴ the assumption that the circle does not pass through PP, must be false, and AA, BB, PP and QQ lie on the circumference of a circle.
- Find the values of the unknown letters.
Cyclic quadrilaterals are quadrilaterals with all four vertices lying on the circumference of a circle. The vertices of a cyclic quadrilateral are said to be concyclic.
Theorem 7 The opposite angles of a cyclic quadrilateral are supplementary.
Proof:
Consider a circle, with centre OO. Draw a cyclic quadrilateral ABPQABPQ. Draw AOAO and POPO.
The aim is to prove that ABP^+AQP^=180∘ABP^+AQP^=180∘ and QAB^+QPB^=180∘QAB^+QPB^=180∘.
O
1
^
=
2
A
B
P
^
∠
's at centre
O
2
^
=
2
A
Q
P
^
∠
's at centre
But,
O
1
^
+
O
2
^
=
360
∘
∴
2
A
B
P
^
+
2
A
Q
P
^
=
360
∘
∴
A
B
P
^
+
A
Q
P
^
=
180
∘
Similarly
,
QAB
^
+
QPB
^
=
180
∘
O
1
^
=
2
A
B
P
^
∠
's at centre
O
2
^
=
2
A
Q
P
^
∠
's at centre
But,
O
1
^
+
O
2
^
=
360
∘
∴
2
A
B
P
^
+
2
A
Q
P
^
=
360
∘
∴
A
B
P
^
+
A
Q
P
^
=
180
∘
Similarly
,
QAB
^
+
QPB
^
=
180
∘
(7)Theorem 8 (Converse of Theorem (Reference)) If the opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic.
Proof:
Consider a quadrilateral ABPQABPQ, such that ABP^+AQP^=180∘ABP^+AQP^=180∘ and QAB^+QPB^=180∘QAB^+QPB^=180∘.
The aim is to prove that points AA, BB, PP and QQ lie on the circumference of a circle.
By contradiction. Assume that point PP does not lie on a circle drawn through points AA, BB and QQ. Let the circle cut APAP (or APAP extended) at point RR. Draw BRBR.
Q
A
B
^
+
Q
R
B
^
=
180
∘
opp.
∠
's of cyclic quad
but
QAB
^
+
QPB
^
=
180
∘
(
given
)
∴
Q
R
B
^
=
Q
P
B
^
but this cannot be true since
QRB
^
=
Q
P
B
^
+
R
B
P
^
(
ext
.
∠
of
▵
)
Q
A
B
^
+
Q
R
B
^
=
180
∘
opp.
∠
's of cyclic quad
but
QAB
^
+
QPB
^
=
180
∘
(
given
)
∴
Q
R
B
^
=
Q
P
B
^
but this cannot be true since
QRB
^
=
Q
P
B
^
+
R
B
P
^
(
ext
.
∠
of
▵
)
(8)∴∴ the assumption that the circle does not pass through PP, must be false, and AA, BB, PP and QQ lie on the circumference of a circle and ABPQABPQ is a cyclic quadrilateral.
- Find the values of the unknown letters.
Theorem 9 Two tangents drawn to a circle from the same point outside the circle are equal in length.
Proof:
Consider a circle, with centre OO. Choose a point PP outside the circle. Draw two tangents to the circle from point PP, that meet the circle at AA and BB. Draw lines OAOA, OBOB and OPOP.
The aim is to prove that AP=BPAP=BP.
In ▵OAP▵OAP and ▵OBP▵OBP,
- OA=OBOA=OB (radii)
- ∠OAP=∠OPB=90∘∠OAP=∠OPB=90∘ (OA⊥APOA⊥AP and OB⊥BPOB⊥BP)
- OPOP is common to both triangles.
▵OAP≡▵OBP▵OAP≡▵OBP (right angle, hypotenuse, side)
∴AP=BP∴AP=BP
- Find the value of the unknown lengths.
Theorem 10 The angle between a tangent and a chord, drawn at the point of contact of the chord, is equal to the angle which the chord subtends in the alternate segment.
Proof:
Consider a circle, with centre OO. Draw a chord ABAB and a tangent SRSR to the circle at point BB. Chord ABAB subtends angles at points PP and QQ on the minor and major arcs, respectively.
Draw a diameter BTBT and join AA to TT.
The aim is to prove that APB^=ABR^APB^=ABR^ and AQB^=ABS^AQB^=ABS^.
First prove that AQB^=ABS^AQB^=ABS^ as this result is needed to prove that APB^=ABR^APB^=ABR^.
A
B
S
^
+
A
B
T
^
=
90
∘
(
TB
⊥
SR
)
B
A
T
^
=
90
∘
(
∠
's at centre
)
∴
A
B
T
^
+
A
T
B
^
=
90
∘
(
sum of angles in
▵
BAT
)
∴
A
B
S
^
=
A
B
T
^
However,
AQB
^
=
A
T
B
^
(
angles subtended by same chord
AB
)
∴
A
Q
B
^
=
A
B
S
^
S
B
Q
^
+
Q
B
R
^
=
180
∘
(
SBT
is a str. line
)
A
P
B
^
+
A
Q
B
^
=
180
∘
(
ABPQ
is a cyclic quad
)
∴
S
B
Q
^
+
Q
B
R
^
=
A
P
B
^
+
A
Q
B
^
AQB
^
=
A
B
S
^
∴
A
P
B
^
=
A
B
R
^
A
B
S
^
+
A
B
T
^
=
90
∘
(
TB
⊥
SR
)
B
A
T
^
=
90
∘
(
∠
's at centre
)
∴
A
B
T
^
+
A
T
B
^
=
90
∘
(
sum of angles in
▵
BAT
)
∴
A
B
S
^
=
A
B
T
^
However,
AQB
^
=
A
T
B
^
(
angles subtended by same chord
AB
)
∴
A
Q
B
^
=
A
B
S
^
S
B
Q
^
+
Q
B
R
^
=
180
∘
(
SBT
is a str. line
)
A
P
B
^
+
A
Q
B
^
=
180
∘
(
ABPQ
is a cyclic quad
)
∴
S
B
Q
^
+
Q
B
R
^
=
A
P
B
^
+
A
Q
B
^
AQB
^
=
A
B
S
^
∴
A
P
B
^
=
A
B
R
^
(9)- Find the values of the unknown letters.
Theorem 11 (Converse of (Reference)) If the angle formed between a line, that is drawn through the end point of a chord, and the chord, is equal to the angle subtended by the chord in the alternate segment, then the line is a tangent to the circle.
Proof:
Consider a circle, with centre OO and chord ABAB. Let line SRSR pass through point BB. Chord ABAB subtends an angle at point QQ such that ABS^=AQB^ABS^=AQB^.
The aim is to prove that SBRSBR is a tangent to the circle.
By contradiction. Assume that SBRSBR is not a tangent to the circle and draw XBYXBY such that XBYXBY is a tangent to the circle.
A
B
X
^
=
A
Q
B
^
(
tan
-
chord theorem
)
However
,
ABS
^
=
A
Q
B
^
(
given
)
∴
A
B
X
^
=
A
B
S
^
But
,
ABX
^
=
A
B
S
^
+
X
B
S
^
can only be true if
,
XBS
^
=
0
A
B
X
^
=
A
Q
B
^
(
tan
-
chord theorem
)
However
,
ABS
^
=
A
Q
B
^
(
given
)
∴
A
B
X
^
=
A
B
S
^
But
,
ABX
^
=
A
B
S
^
+
X
B
S
^
can only be true if
,
XBS
^
=
0
(10)If XBS^XBS^ is zero, then both XBYXBY and SBRSBR coincide and SBRSBR is a tangent to the circle.
- Show that Theorem(Reference) also applies to the following two cases:
BDBD is a tangent to the circle with centre
OO.
BO⊥ADBO⊥AD.
Prove that:
- CFOECFOE is a cyclic quadrilateral
- FB=BCFB=BC
- ▵COE///▵CBF▵COE///▵CBF
- CD2=ED.ADCD2=ED.AD
- OEBC=CDCOOEBC=CDCO
- Step 1. To show a quadrilateral is cyclic, we need a pair of opposite angles to be supplementary, so let's look for that. :
F
O
E
^
=
90
∘
(
BO
⊥
OD
)
F
C
E
^
=
90
∘
(
∠
subtended by diameter
AE
)
∴
C
F
O
E
is a cyclic quad
(
opposite
∠
's supplementary
)
F
O
E
^
=
90
∘
(
BO
⊥
OD
)
F
C
E
^
=
90
∘
(
∠
subtended by diameter
AE
)
∴
C
F
O
E
is a cyclic quad
(
opposite
∠
's supplementary
)
(11) - Step 2. Since these two sides are part of a triangle, we are proving that triangle to be isosceles. The easiest way is to show the angles opposite to those sides to be equal. :
Let OEC^=xOEC^=x.
∴
F
C
B
^
=
x
(
∠
between tangent
BD
and chord
CE
)
∴
B
F
C
^
=
x
(
exterior
∠
to cyclic quad
CFOE
)
∴
B
F
=
B
C
(
sides opposite equal
∠
's in isosceles
▵
BFC
)
∴
F
C
B
^
=
x
(
∠
between tangent
BD
and chord
CE
)
∴
B
F
C
^
=
x
(
exterior
∠
to cyclic quad
CFOE
)
∴
B
F
=
B
C
(
sides opposite equal
∠
's in isosceles
▵
BFC
)
(12) - Step 3. To show these two triangles similar, we will need 3 equal angles. We already have 3 of the 6 needed angles from the previous question. We need only find the missing 3 angles. :
C
B
F
^
=
180
∘
-
2
x
(
sum of
∠
's in
▵
BFC
)
O
C
=
O
E
(
radii of circle
O
)
∴
E
C
O
^
=
x
(
isosceles
▵
COE
)
∴
C
O
E
^
=
180
∘
-
2
x
(
sum of
∠
's in
▵
COE
)
C
B
F
^
=
180
∘
-
2
x
(
sum of
∠
's in
▵
BFC
)
O
C
=
O
E
(
radii of circle
O
)
∴
E
C
O
^
=
x
(
isosceles
▵
COE
)
∴
C
O
E
^
=
180
∘
-
2
x
(
sum of
∠
's in
▵
COE
)
(13)-
C
O
E
^
=
C
B
F
^
C
O
E
^
=
C
B
F
^
-
E
C
O
^
=
F
C
B
^
E
C
O
^
=
F
C
B
^
-
O
E
C
^
=
C
F
B
^
O
E
C
^
=
C
F
B
^
∴
▵
C
O
E
|||
▵
C
B
F
(
3
∠
's equal
)
∴
▵
C
O
E
|||
▵
C
B
F
(
3
∠
's equal
)
(14) - Step 4.
- Step 1. This relation reminds us of a proportionality relation between similar triangles. So investigate which triangles contain these sides and prove them similar. In this case 3 equal angles works well. Start with one triangle. :
In ▵EDC▵EDC
C
E
D
^
=
180
∘
-
x
(
∠
's on a str. line
AD
)
E
C
D
^
=
90
∘
-
x
(
complementary
∠
's
)
C
E
D
^
=
180
∘
-
x
(
∠
's on a str. line
AD
)
E
C
D
^
=
90
∘
-
x
(
complementary
∠
's
)
(15) - Step 2. Now look at the angles in the other triangle. :
In ▵ADC▵ADC
A
C
E
^
=
180
∘
-
x
(
sum of
∠
's
ACE
^
and
ECO
^
)
C
A
D
^
=
90
∘
-
x
(
sum of
∠
's in
▵
CAE
)
A
C
E
^
=
180
∘
-
x
(
sum of
∠
's
ACE
^
and
ECO
^
)
C
A
D
^
=
90
∘
-
x
(
sum of
∠
's in
▵
CAE
)
(16) - Step 3. The third equal angle is an angle both triangles have in common. :
Lastly, ADC^=EDC^ADC^=EDC^ since they are the same ∠∠.
- Step 4. Step 4d: Now we know that the triangles are similar and can use the proportionality relation accordingly. :
∴
▵
A
D
C
|||
▵
C
D
E
(
3
∠
's equal
)
∴
E
D
C
D
=
C
D
A
D
∴
C
D
2
=
E
D
.
A
D
∴
▵
A
D
C
|||
▵
C
D
E
(
3
∠
's equal
)
∴
E
D
C
D
=
C
D
A
D
∴
C
D
2
=
E
D
.
A
D
(17)
- Step 5.
- Step 1.
This looks like another proportionality relation with a little twist, since not all sides are contained in 2 triangles. There is a quick observation we can make about the odd side out, OEOE. :
O
E
=
C
D
(
▵
OEC
is isosceles
)
O
E
=
C
D
(
▵
OEC
is isosceles
)
(18) - Step 2. With this observation we can limit ourselves to proving triangles BOCBOC and ODCODC similar. Start in one of the triangles. :
In ▵BCO▵BCO
O
C
B
^
=
90
∘
(
radius
OC
on tangent
BD
)
C
B
O
^
=
180
∘
-
2
x
(
sum of
∠
's in
▵
BFC
)
O
C
B
^
=
90
∘
(
radius
OC
on tangent
BD
)
C
B
O
^
=
180
∘
-
2
x
(
sum of
∠
's in
▵
BFC
)
(19) - Step 3. Then we move on to the other one. :
In ▵OCD▵OCD
O
C
D
^
=
90
∘
(
radius
OC
on tangent
BD
)
C
O
D
^
=
180
∘
-
2
x
(
sum of
∠
's in
▵
OCE
)
O
C
D
^
=
90
∘
(
radius
OC
on tangent
BD
)
C
O
D
^
=
180
∘
-
2
x
(
sum of
∠
's in
▵
OCE
)
(20) - Step 4. Again we have a common element. :
Lastly, OCOC is a common side to both ▵▵'s.
- Step 5. Step 5e: Then, once we've shown similarity, we use the proportionality relation , as well as our first observation, appropriately. :
∴
▵
B
O
C
|||
▵
O
D
C
(
common side and 2 equal angles
)
∴
C
O
B
C
=
C
D
C
O
∴
O
E
B
C
=
C
D
C
O
(
OE
=
CD
isosceles
▵
OEC
)
∴
▵
B
O
C
|||
▵
O
D
C
(
common side and 2 equal angles
)
∴
C
O
B
C
=
C
D
C
O
∴
O
E
B
C
=
C
D
C
O
(
OE
=
CD
isosceles
▵
OEC
)
(21)
FDFD is drawn parallel to the tangent
CBCB
Prove that:
- FADEFADE is cyclic
- ▵AFE|||▵CBD▵AFE|||▵CBD
- FC.AGGH=DC.FEBDFC.AGGH=DC.FEBD
- Step 1. In this case, the best way to show FADEFADE is a cyclic quadrilateral is to look for equal angles, subtended by the same chord. :
Let ∠BCD=x∠BCD=x
∴
∠
C
A
H
=
x
(
∠
between tangent BC and chord CE
)
∴
∠
F
D
C
=
x
(
alternate
∠
,
FD
∥
CB
)
∴
FADE
is a cyclic quad
(
chord FE subtends equal
∠
's
)
∴
∠
C
A
H
=
x
(
∠
between tangent BC and chord CE
)
∴
∠
F
D
C
=
x
(
alternate
∠
,
FD
∥
CB
)
∴
FADE
is a cyclic quad
(
chord FE subtends equal
∠
's
)
(22) - Step 2.
- Step 1.
To show these 2 triangles similar we will need 3 equal angles. We can use the result from the previous question. :
Let ∠FEA=y∠FEA=y
∴
∠
F
D
A
=
y
(
∠
's subtended by same chord
AF
in cyclic quad
FADE
)
∴
∠
C
B
D
=
y
(
corresponding
∠
's,
FD
∥
CB
)
∴
∠
F
E
A
=
∠
C
B
D
∴
∠
F
D
A
=
y
(
∠
's subtended by same chord
AF
in cyclic quad
FADE
)
∴
∠
C
B
D
=
y
(
corresponding
∠
's,
FD
∥
CB
)
∴
∠
F
E
A
=
∠
C
B
D
(23) - Step 2. We have already proved 1 pair of angles equal in the previous question. :
∠
B
C
D
=
∠
F
A
E
(
above
)
∠
B
C
D
=
∠
F
A
E
(
above
)
(24) - Step 3. Proving the last set of angles equal is simply a matter of adding up the angles in the triangles. Then we have proved similarity. :
∠
A
F
E
=
180
∘
-
x
-
y
(
∠
's in
▵
AFE
)
∠
C
B
D
=
180
∘
-
x
-
y
(
∠
's in
▵
CBD
)
∴
▵
A
F
E
|||
▵
C
B
D
(
3
∠
's equal
)
∠
A
F
E
=
180
∘
-
x
-
y
(
∠
's in
▵
AFE
)
∠
C
B
D
=
180
∘
-
x
-
y
(
∠
's in
▵
CBD
)
∴
▵
A
F
E
|||
▵
C
B
D
(
3
∠
's equal
)
(25)
- Step 3.
- Step 1. This equation looks like it has to do with proportionality relation of similar triangles. We already showed triangles AFEAFE and CBDCBD similar in the previous question. So lets start there. :
D
C
B
D
=
F
A
F
E
∴
D
C
.
F
E
B
D
=
F
A
D
C
B
D
=
F
A
F
E
∴
D
C
.
F
E
B
D
=
F
A
(26) - Step 2. Now we need to look for a hint about side FAFA. Looking at triangle CAHCAH we see that there is a line FGFG intersecting it parallel to base CHCH. This gives us another proportionality relation. :
A
G
G
H
=
F
A
F
C
(
FG
∥
CH
splits up lines
AH
and
AC
proportionally
)
∴
F
A
=
F
C
.
A
G
G
H
A
G
G
H
=
F
A
F
C
(
FG
∥
CH
splits up lines
AH
and
AC
proportionally
)
∴
F
A
=
F
C
.
A
G
G
H
(27) - Step 3. We have 2 expressions for the side FAFA. :
∴
F
C
.
A
G
G
H
=
D
C
.
F
E
B
D
∴
F
C
.
A
G
G
H
=
D
C
.
F
E
B
D
(28)
