Transcript H11 Alternate Segment Theorem - Chartwell

Teach GCSE Maths
Alternate Segment Theorem
Teach GCSE Maths
Alternate Segment
Theorem
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© Christine Crisp
When we draw a chord joining 2 points on the
circumference of a circle, we form 2 segments.
major segment
O
minor segment
The major segment is the larger one.
If we now draw a tangent at one end of the chord
. . .
we make an acute angle
P
between the tangent
x
and the chord, TAB .
If we also draw an angle
in the major segment . . .
the alternate segment
theorem says
TAB = APB
O
A
B
x
T
• P is formed by lines from A and B.
• P can be anywhere on the circumference of the
major segment.
If we now draw a tangent at one end of the chord
. . .
we make an acute angle
P
between the tangent
x

and the chord, TAB .
P
x
If we also draw an angle
O
B
in the major segment . . .
the alternate segment
theorem says
x
A
TAB = APB
T
• P is formed by lines from A and B.
• P can be anywhere on the circumference of the
major segment.
The same is true for the obtuse angle but it is
equal to the angle in the minor segment.
This property of angles
is called the
alternate segment
theorem.
The word “alternate” is
used because the equal
angles are on different
sides of the chord.
O
y
y
We’ll use an example to see why the alternate
segment theorem is true.
Each time the animation pauses, decide with your
We want
to the
showreason
that angle
= 50 angle.
partner
what
is forADB
the given

TAC = 90
e.g.
D
C
Angle between
50
tangent and radius
50

CAB = 40
90 - 50
B 
ABC = 90
O
A
40
50

T

Angle in a
semi-circle
ACB = 50
3rd angle of triangle
ADB = 50 Angles in the same segment
To prove the alternate segment theorem we use
the same steps as in the example.
D
C
O
A
Let  TAB = x
 TAC = 90
B
a
x
T
( angle between
tangent and radius )
 CAB = a = 90 - x
 CBA = 90
( angle in a semi-circle )
 ACB = 180 – 90 – a ( 3rd angle of triangle )
To prove the alternate segment theorem we use
the same steps as in the example.
D
C
O
A
Let  TAB = x
 TAC = 90
B
a
x
T
( angle between
tangent and radius )
 CAB = a = 90 - x
 CBA = 90
( angle in a semi-circle )
 ACB = 180 – 90 – a ( 3rd angle of triangle )
= 90 – ( 90 - x )
To prove the alternate segment theorem we use
the same steps as in the example.
D
x
C
x
O
A
Let  TAB = x
 TAC = 90
B
a
x
T
( angle between
tangent and radius )
 CAB = a = 90 - x
 CBA = 90
( angle in a semi-circle )
 ACB = 180 – 90 – a ( 3rd angle of triangle )
= 90 – ( 90 - x )
= 90 – 90 + x = x

ADB = x ( angle in the same segment )
To prove the alternate segment theorem we use
the same steps as in the example.
D
x
Let  TAB = x
 TAC = 90
O
A
B
x
T
( angle between
tangent and radius )
 CAB = a = 90 - x
 CBA = 90
( angle in a semi-circle )
 ACB = 180 – 90 – a ( 3rd angle of triangle )
= 90 – ( 90 - x )
= 90 – 90 + x = x

ADB = x ( angle in the same segment )
SUMMARY
The theorems involving tangents are:
•
The angle between a tangent and the radius at
the point of contact is always 90.
•
The tangents from an external point are equal
in length.
•
The angle between a tangent and chord equals
the angle in the alternate segment.
SUMMARY
When solving problems we might also use the
following:
•
The perpendicular from the centre to a chord
bisects the chord.
•
The angle at the centre is twice the angle at the
circumference.
•
Angles in the same segment are equal.
•
The angle in a semi-circle is always 90.
•
The sum of the opposite angles of any cyclic
quadrilateral is 180.
e.g. In the following, the red line is a tangent. Find
a and b giving the reasons.
a
b
Solution:
115
a = 115
(alternate segment )
b = 2a = 230 ( angle at the centre = twice the
angle at the circumference )
EXERCISE
1. In the following, O is the centre of the circle
and TB is a tangent. Find x, y and z giving the
reasons.
A
x
65
x = 65 ( angle in the
alternate segment )
C
z
25
O
y
130
65
T
Solution:
B
y = 130
( angle at the centre = twice
the angle at the circumference )
z = 25 ( isosceles triangle:
OC and OB are radii )
EXERCISE
2. In the following, find angles TBA, BCA and
ADB giving the reasons.
T
65 B
50
Solution:
D 115
O
65
TA = TB
( tangents from one point )
A
 TBA = 65 ( triangle TAB is isosceles )
 BCA = 65 ( angle in the alternate segment )
 ADB = 115 ( opposite angles of cyclic quad. )
C