D CIRCLE GEOMETRY CIRCLE TERMINOLOGY

Transcription

145
GEOMETRY (Chapter 2)
D
CIRCLE GEOMETRY
CIRCLE TERMINOLOGY
If an arc is less than half the circle, it is called
a minor arc. If an arc is greater than half the
circle, it is called a major arc.
C
B
C
B
A chord divides the interior of a circle into two
regions called segments. The larger region is
called a major segment and the smaller region
is called a minor segment.
major arc BC
minor arc BC
A
Consider minor arc BC.
We can say that the arc BC subtends the angle
BAC at A which lies on the circle.
O
E
We also say that the arc BC subtends an angle at
the centre of the circle, which is angle BOC.
major segment
B
C
SA
M
PL
minor segment
CIRCLE THEOREMS
Name of theorem
Statement
Diagram
Angle in a
semi-circle
The angle in a semi-circle
is a right angle.
C
A
Proof:
O
b = 90±
ACB
GEOMETRY
PACKAGE
B
Since OA = OB = OC, triangles OAC and OBC are isosceles.
) ®1 = ®2 and ¯ 1 = ¯ 2
C
Now in triangle ABC,
®1 + ¯ 1 + (®2 + ¯ 2 ) = 180±
) 2® + 2¯ = 180±
) ® + ¯ = 90±
®2 ¯2
¯1
®1
A
O
fisosceles triangleg
B
fangles of a triangleg
b is a right angle.
) ACB
Converse 1:
B
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A
If M is the midpoint of the hypotenuse of a right angled triangle,
then a circle can be drawn through A, B, and C with M its centre on
diameter [AC].
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Further Mathematics
146
Converse 2:
B
b is a right angle, then [AC] is
If A, B, and C lie on a circle and ABC
a diameter of the circle.
C
A
Proof of converse 2:
Let [AD] be a diameter of the circle with centre M. Join [BD].
B
b is a right angle fangle in a semi-circleg
) ABD
A
b C is a right angle.
Now C also lies on the circle such that AB
C
D
b = 90± ¡ 90± = 0
) CBD
) C and D are coincident
) [AC] is a diameter of the circle.
SA
M
PL
E
M
Name of theorem
Statement
Diagram
Chord of a
circle
The perpendicular from
the centre of a circle to a
chord, bisects the chord.
AM = BM
A
O
GEOMETRY
PACKAGE
M
B
Proof:
A
OA = OB
fequal radiig
) triangle OAB is isosceles
) AM = MB fisosceles triangleg
O
M
B
Converse 1:
The line from the centre of a circle to the midpoint of a chord, is perpendicular to the chord.
Converse 2:
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The perpendicular bisector of a chord of a circle, passes through the circle’s centre.
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Further Mathematics
147
Proof of converse 2:
Let X be any point on the perpendicular bisector of [AB].
4s XAM and XBM are congruent fSASg
P
) XA = XB
X
Now choose X so that XA = XB = r, where r is the radius of
the circle.
A
M
) X is necessarily the circle’s centre.
fdistance r from both A and B, and lies within the circleg
B
) the perpendicular bisector of the chord passes through the
circle’s centre.
Diagram
Radius-tangent
The tangent to a circle
is perpendicular to the
radius at the point of
contact.
b = 90±
OAT
E
Statement
GEOMETRY
PACKAGE
O
SA
M
PL
Name of theorem
A
T
Proof:
Consider a circle with centre O, and a tangent to the circle with point
of contact A.
Suppose P is any point on the tangent and P is not at A.
O
) P lies outside the circle.
A
) OA is the shortest distance from O to the tangent.
T
P
) [OA] is perpendicular to the tangent.
Name of theorem
Statement
Diagram
Angle at the
centre
The angle at the centre of
a circle is twice the angle
on the circle subtended by
the same arc.
b = 2ACB
b
AOB
C
GEOMETRY
PACKAGE
O
B
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Further Mathematics
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Proof:
OA = OC = OB fequal radiig
C
®2
) triangles AOC and OBC are isosceles
¯1
) ®1 = ®2 and ¯ 1 = ¯ 2
¯2
O
®1 2®
B
2¯
b = 2® and BOX
b = 2¯
But AOX
fexterior angle of a triangleg
b = 2® + 2¯
) AOB
b
= 2 £ ACB
A
X
The following diagrams show
other cases of the angle at the
centre theorem. These cases
can be easily shown using the
geometry package.
(1)
(2)
O
2®
(3)
2®
O
E
®
O
GEOMETRY
PACKAGE
®
2®
SA
M
PL
®
In case (2), letting 2® = 180± we have another proof of the angle in a semi-circle theorem. So, the
angle in a semi-circle could be considered as a corollary of the angle at the centre theorem.
Corollary
Statement
Diagram
Angles subtended by an
arc on the circle are
equal in size.
Angles subtended
by the same arc
b
b = ACB
ADB
D
C
A
GEOMETRY
PACKAGE
B
Proof:
D
bB = 1 ®
AD
fangle at the centreg
2
1
b = ®
and ACB
fangle at the centreg
2
b
b = ACB
) ADB
C
O
®
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Further Mathematics
Name of theorem
Statement
Tangents from
an external point
Tangents from an
external point are
equal in length, and
the line joining the
point to the centre
bisects the angle at
the point.
149
Diagram
AP = BP
PO = Bb
PO
Ab
A
O
GEOMETRY
PACKAGE
P
B
Proof:
We observe that:
b = OB
b P = 90± fradius-tangentg
² OAP
A
² OA = OB
P
fequal radiig
² OP is common to both
E
O
) 4s OAP and OBP are congruent (RHS).
B
SA
M
PL
Consequently, AP = BP and Ab
PO = Bb
PO.
Name of theorem
Statement
Angle between
tangent and chord
The angle between
a tangent and a
chord at the point of
contact, is equal to
the angle subtended
by the chord in the
alternate segment.
Proof:
Diagram
C
T
A
b = BCA
b
BAS
GEOMETRY
PACKAGE
B
S
We draw AOX and BX.
b = 90±
XAS
b X = 90±
AB
X
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So, in 4ABX,
b = 180± ¡ 90± ¡ (90± ¡ ®) = ®
BXA
b
b = BCA
But BXA
fangles subtended by the same arcg
b = BAS
b =®
) BCA
®
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T
fradius-tangentg
fangle in a semi-circleg
b =®
Let BAS
b = 90± ¡ ®
) BAX
B
O
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C
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Further Mathematics
150
Name of theorem
Statement
Diagram
Intersecting
circles
The line joining
the centres of two
intersecting circles
bisects the common
chord at right angles.
P
M
X
GEOMETRY
PACKAGE
Y
Q
[XY] ? [PQ]
MP = MQ
USING CIRCLE THEOREMS
b C is a right angle:
Show that AD
E
Example 6
B
C
SA
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A
D
Since AB = BD, 4ABD is isosceles.
B
A ®1
) ®1 = ®2
C
¯2
Likewise, ¯ 1 = ¯ 2 in isosceles triangle BCD.
®2 ¯1
Thus in triangle ADC,
® + (® + ¯) + ¯ = 180±
fangles of a triangleg
) 2® + 2¯ = 180±
) ® + ¯ = 90±
D
b C is a right angle.
) AD
Alternatively:
Since BA = BC = BD, a circle with centre B can be drawn through A, D, and C.
[AC] is a diameter.
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b C is a right angle. fangle in a semi-circleg
) AD
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Further Mathematics
151
Example 7
C
Given a circle with centre O, and a point A on the circle,
a smaller circle with diameter [OA] is drawn. [AC] is
any line drawn from A to the larger circle, cutting the
smaller circle at B.
B
O
A
Prove that the smaller circle will always bisect [AC].
C
Join [OA], [OC], and [OB].
b A is a right angle.
Now OB
fangle in a semi-circleg
B
A
Thus [OB] is the perpendicular from the centre of the
circle to the chord [AC].
E
O
) [OB] bisects [AC]. fchord of a circle theoremg
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Thus B always bisects [AC].
EXERCISE 2D
1 O is the centre of two concentric circles. [AB] is a tangent to the
smaller circle at C. A and B are both on the larger circle. Prove
that AC = BC.
O
A
C
B
2 Triangle PQR is inscribed in a circle. The angle bisector of
PR meets [QR] at S, and the circle at T.
Qb
If PQR is inscribed
Prove that PQ:PR = PS:PT.
in a circle, a circle is
drawn through its
three vertices.
3
A
O is the centre of two concentric circles.
[AB] is a diameter of the smaller circle.
Tangents at A and B are drawn to cut the larger circle at
M and N respectively.
Prove that AM = BN.
M
O
N
B
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4 The tangent at P to a circle meets the chord [QR] produced at the point S. Prove that triangles SPQ
and SRP are similar.
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5 P, Q, R, and S are distinct points on a circle, and are in cyclic order. The diagonals of PQRS meet
at A. Prove that triangles PQA and SRA are similar.
6 Prove the ‘intersecting circles’ theorem.
P
7 Triangle PQR is isosceles with PQ = PR. A semi-circle
with diameter [PR] is drawn which cuts [QR] at X.
Prove that X is the midpoint of [QR].
O
Q
R
X
Y
8 [AB] is a diameter of a circle with centre O. X is a
point on the circle, and [AX] is produced to Y such that
OX = XY.
b is three times the size of XOY.
b
Prove that YOB
E
X
B
O
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A
9 Triangle PQR is isosceles with PQ = QR. PQR is inscribed in a circle. [XP] is a tangent to the
circle. Prove that [QP] bisects angle XPR.
10 [AB] is a diameter of a circle with centre O. [CD] is a chord parallel to [AB]. Prove that [BC]
bisects the angle DCO, regardless of where [CD] is located.
b and QOR
b are
11 [PQ] and [RS] are two perpendicular chords of a circle with centre O. Prove that POS
supplementary.
b of 4XYZ meets [YZ] at W. When a circle is drawn through X, it touches
12 The bisector of YXZ
b = ZWQ.
b
[YZ] at W, and cuts [XY] and [XZ] at P and Q respectively. Prove that YWP
b cuts [BC] at P, and the circle at Q.
13 A, B, and C are three points on a circle. The bisector of CAB
b Q.
PC = AB
Prove that Ab
14 [AB] and [DC] are parallel chords of a circle. [AC] and [BD] intersect at E. Prove that:
a triangles ABE and CDE are isosceles
b AC = BD.
15 P is any point on a circle. [QR] is a chord of the circle parallel to the tangent at P. Prove that
triangle PQR is isosceles.
16 Two circles intersect at A and B.
[AX] and [AY] are diameters, as shown.
Prove that X, B, and Y are collinear.
A
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Further Mathematics
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P
17 Two circles intersect at A and B. Straight lines [PQ]
and [XY] are drawn through A to meet the circles
as shown.
b P = YB
b Q.
Show that XB
Y
A
X
Q
B
18 Triangle PQR is inscribed in a circle with [PR] as a diameter. The perpendicular from P to the
tangent at Q, meets the tangent at S. Prove that [PQ] bisects angle SPR.
19 Tangents are drawn from a fixed point C to a fixed
circle, meeting it at A and B. [XY] is a moving tangent
which meets [AC] at X, and [BC] at Y. Prove that
triangle XYC has constant perimeter.
A
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E
X
20 [AB] is a diameter of a circle. The tangent at X cuts
the diameter produced at Y. [XZ] is perpendicular to
[AY] at Z on [AY]. Prove that [XB] and [XA] are the
b and ZXT
b respectively.
bisectors of ZXY
C
T
X
A
Z
21 In the given figure, AF = FC and PE = EC.
a Prove that triangle FPA is isosceles.
P
b Prove that AB + BE = EC.
Y
B
Y
B
F
B
E
C
A
22 Tangents from the external points P, Q, R, and S
form a quadrilateral. This is called a circumscribed
polygon.
What can be deduced about the opposite sides of
the circumscribed quadrilateral?
Prove your conjecture.
Q
B
P
C
A
S
D
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23 [POQ] is a diameter of a circle with centre O, and R is any other point on the circle. The tangent
b is a right angle.
at R meets the tangents at P and Q at S and T respectively. Show that SOT
24 [PQ] and [PR] are tangents from an external point P to a circle with centre O.
[PS] is perpendicular to [PQ] and meets [OR] produced at S.
[QR] produced meets [PS] produced at T. Show that triangle STR is isosceles.
y
25 A solid thin bar [AB] moves so that A remains on
the x-axis and B remains on the y-axis. There is a
small light source at P, the midpoint of [AB].
Without using coordinate geometry methods, prove
that as A and B move to all possible positions, the
light traces out a circle.
DEMO
B
P
x
A
26
(AB) is a common tangent to two circles. Prove that:
E
a the tangent through the point of contact C bisects
[AB]
C
b is a right angle.
b ACB
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B
A
27 Two circles touch externally at B. (CD) is a common
tangent touching the circles at D and C.
[DA] is a diameter.
Prove that A, B, and C are collinear.
D
C
B
A
28
For the given figure, prove that
QP2 = QA:QB.
B
A
Q
S
P
R
29 Two circles touch internally at point A. Chord [AC]
of the larger circle cuts the smaller circle at B, and
chord [AE] cuts the smaller circle at D.
Prove that [BD] is parallel to [CE].
C
B
A
D
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30 Two circles touch internally at point P. The tangent to the inner circle at Q meets the outer circle
PS.
at R and S. Prove that [QP] bisects Rb
31 A and B are the goalposts on a football field. A
photographer wants to find the point P on the boundary
PB,
line such that his viewing angle of the goal, Ab
is maximised. Prove that P should be chosen so the
boundary line is a tangent to the circle through A, B,
and P.
A
B
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Further Mathematics

D CIRCLE GEOMETRY CIRCLE TERMINOLOGY

Transcription

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