Download Circles in Euclidean Geometry, Part II

Section 4.5: Circles in Euclidean Geometry (Part 2: After Corollary 4.5.13)
Theorem 4.5.11, The Inscribed Angle Theorem : The measure of an angle
inscribed in an arc is one-half the measure of its intercepted arc.
Corollary 4.5.12: An angle inscribed in a semicircle is a right angle.
Theorem (NIB) 4.6, The Hypotenuse Diameter Theorem:
For any right triangle, the hypotenuse is the diameter of a circle containing the
right-angle vertex.
That is, if AB is a diameter of circle C(O, OA) and if APB is a right triangle
with hypotenuse AB , then the vertex P is a point on the circle C(O, OA) which has
the hypotenuse as its diameter
Proof: Let APB be a right triangle with right angle at vertex P and hypotenuse AB .
Let O be the midpoint of the hypotenuse AB ; thus, C(O, OA) is the circle which has the
hypotenuse AB as diameter.
Assume, by way of contradiction, that P is not on the circle C(O, OA) with the hypotenuse AB
as diameter.
Case 1: If AP is tangent to the circle, then  PAB is also a right angle
and APB has two right angles, which makes its angle sum greater than
180º. This contradicts the fact that its angle sum is equal to 180º.
P
B
O
A
Case 2: If AP is not tangent to the circle, then either P is outside the
circle or P is inside the circle.
Suppose P is outside the circle; then, AP will intersect the circle, say
at point Q with A – Q – P or with Q – A – P. In both cases,  AQB is a
right angle by Corollary 4.5.12, and so BQP has two right angles. Thus,
BQP has an angle sum greater than 180º . This contradicts the fact that its
angle sum is equal to 180º.
P
Q
B
O
A
Suppose P is inside the circle; then, we can extend AP until it intersects
the circle, say at point Q with A – P – Q . By Corollary 4.5.12,  AQB is
a right angle and so BQP has two right angles. Thus, BQP has an angle
sum greater than 180º. This contradicts the fact that its angle sum is equal to
180º.
If P is not on the circle, then all possible cases lead to a contradiction.
Thus, P is a point on the circle.
QED
Q
P
B
O
A
2
Definitions:
Four (or more) points are said to be concyclic if there exists a circle containing all of them
together.
A quadrilateral is said to be a cyclic quadrilateral if the quadrilateral can be circumscribed.
Thus, a quadrilateral is a cyclic quadrilateral if and only if its four vertices are concyclic.
In these terms, Theorem (NIB) 4.8 below states that a quadrilateral is a cyclic quadrilateral
if and only if its opposite angles are supplementary.
Theorem (NIB) 4.7, The Common Hypotenuse Theorem:
If  ABF and  ABG are two right triangles with segment AB as a common hypotenuse,
then the circle which has the common hypotenuse AB as its diameter will contain the points F
and G, the “right-angle vertices” of the two right triangles. Consequently, all four points A, B, F
and G will be concyclic.
Proof: Let O be the midpoint of segment AB and consider circle C(O, OA), the circle with
AB as diameter.
By the Hypotenuse Diameter Theorem (Theorem (NIB) 4.6),
since vertex F is the right-angle vertex in  ABF, F must be a
point on the circle with AB as diameter, C(O, OA).
F
Similarly, by the Hypotenuse Diameter Theorem (Theorem
(NIB) 4.6), since vertex G is the right-angle vertex in  ABG,
G must be a point on the circle with AB as diameter, C(O, OA).
G
A
O
B
Thus, the circle with AB as a diameter contains all four points
A, B, F and G.
Therefore, the four points are concyclic.
QED
One situation in which The Common Hypotenuse
Theorem always applies is when we consider two
vertices of a triangle and the feet of the two
segment altitudes drawn from those two vertices.
This theorem will show that there will always
exist a circle containing these four points (or three
if two coincide), namely, the two vertices and the
two feet of the altitudes.
C
B
FA
O
FB
A
3
Theorem (NIB) 4.8, The Circumscribable Quadrilateral Theorem :
Quadrilateral ABCD can be circumscribed if and only if its opposite angles are
supplementary.
Proof:
Part I:
Suppose quadrilateral ABCD can be circumscribed. Suppose that circle C(O, OA) is a circle
passing through A, B, C, and D.
Draw chord AC.
B
m(  B ) = ½ m( arc ADC ) by the Inscribed Angle Thm.
m(  D ) = ½ m( arc ABC ) by the same theorem.
m(  B ) + m(  D ) =
C
A
½ (m( arc ADC ) + m( arc ABC ) ) =
O
½ ( 360º) = 180º .
Therefore,  B and  D are supplementary.
Similarly,  A and  C are supplementary.
Therefore, opposite angles in quadrilateral ABCD are
supplementary.
D
Part II:
Suppose that opposite angles in quadrilateral ABCD are supplementary.
There is a unique circle, C(O, OA) passing through the three points
E
A, C, and D. Suppose (by way of contradiction) that point B is not
B
on this circle. Suppose that B is inside the circle (the proof when B
is outside is essentially the same). Extend segment AB to segment
A
AE where E is on the circle. By Part I,  E and  D are
C
O
supplementary. Thus,  E and  B are congruent since both are
supplementary to  D . But, since  B is an exterior angle of
D
 BEC , m( B ) > m( E ) by the Exterior Angle Theorem, a contradiction.
Thus, point B is on the circle, so quadrilateral ABCD has been circumscribed.
Recall that it was pointed out that, if B were outside of the circle, the proof is essentially the
same. QED
4
Theorems 4.5.14 – 4.5.18 are “Angle Theorems” because they relate the angle
formed by chords, tangents and secants with the measures of the arcs of the circle
that they intercept. Thus, they all have the word “Angle” in their names.
Theorems 4.5.20 – 4.5.22 are “Segment Product Theorems” because they show
that the products of the lengths of certain segments are equal. They all have the
word “Segment” in their names.
Corollary 4.5.19 is “Segment Theorem” but it is not a “Segment Product” therem.
Q
Definitions:
The phrase “tangent segment” refers to the
segment of a tangent line of a circle which
connects a point in the exterior of the circle to the
point of tangency on the circle.
P
Segment PQ is a "tangent segment".
The phrase “secant segment” refers to either
one of the two segments of a secant line of a
circle which connects a point in the exterior of
the circle to one of the two points of
intersection of the secant line and the circle.
B
A
P
The phrase “chord segment” refers to either one of
the two segments of a chord of a circle which
connects a point on the chord in the interior of the
circle to one of the two endpoints of the chord on
the circle.
Segments PA and PB
are "secant segments".
A
P
B
Segments PA and PB
are "chord segments".
5
Theorem 4.5.14, The Two-chord Angle Theorem: When two chords intersect
inside a circle, the measure of the angle formed by two chords is the average of the
measures of the arcs intercepted by the angle and its vertical angle.
In the figures below,
1
2
m(  AEB ) 
x  y
.
A
A
D
4
x
y
D
E
5
Q 2
y
B
C
3
x
P
1
E
B
C
m(  1 ) = m(  3 ) + m(  4 ) by Corollary 4.2.3, the Euclidean Exterior Angle Theorem.
m(  3 ) = (1/2) m( APB ) = (1/2) x
m(  4 ) = (1/2) m( DQC ) = (1/2) y
and
by the Inscribed Angle Theorem.
m(  1 )
=
(1/2) x + (1/2) y
= (1/2) ( x + y ) = (1/2) | x + y | .
Theorem 4.5.15: The Two-Secant Angle Theorem: If two secants intersect at a
point in the exterior of a circle, the measure of the angle at the point of intersection
is ½ the positive difference of the two intercepted arcs.
In the figures below,
m(  APB ) 
1
2
x  y
.
Note: The arc further from P has greater measure. Here, x > y .
Thus, (1/2) | x -- y | =
(1/2) ( x -- y ) since ( x -- y ) > 0 .
A
A
2
x
y
B
P
P
1
y
x
3
C
B
m(  3 ) = m(  1 ) + m(  2 ) by Corollary 4.2.3, the Euclidean Exterior Angle Theorem.
m(  3 ) = (1/2) x
m(  1 )
=
and
m(  2 ) = (1/2) y
m(  3 ) -- m(  2 ) =
by the Inscribed Angle Theorem.
(1/2) x -- (1/2) y
= (1/2) ( x -- y ) =
(1/2) | x -- y | .
6
Theorem 4.5.16, The Tangent-Chord Angle Theorem: If AB is tangent to
C(O, OA) at a point A and if AC is a chord such that m( arc APC ) = x , then
1
m(  BAC ) 
x .
2
1
x .
2
m(  BAC ) 
In the figures below, line t is tangent at A and
t
t
B
B
A
A
x
1
2
x
P
P
O
C
C
y
D
AD is a diameter of the circle ,  BAC
m( BAD ) = 90 .
=
= 180 .  (1/2) x + (1/2) y = 90 .
= 90 -- (1/2) y .
m(  2 ) = (1/2) y
m(  1 )
= 2 .
 m(  1 ) + m(  2 ) = 90 .  m(  1 ) = 90 -- m(  2 ) .
m( ACD ) = 180 so x + y
 (1/2) x
=  1 , and  CAD
by the Inscribed Angle Theorem.
90 -- m(  2 )
=
90 -- (1/2) y
=
(1/2) x , by substitution.
Theorem 4.5.17, The Tangent-Secant Angle Theorem: If AB is tangent to
C(O, OB) at a point B and if AD is a secant line to C(O, OB), then m(  BAD ) is
half the positive difference of the two intercepted arcs.
In the figures below,
m(  BAD ) 
B
1
2
x  y
.
B
x
4
y
A
x
3
2
D
A
D
1
The proof is left as an exercise. It is problem #26 in Exercises 4.5.
The proof uses methods similar to those used above.
7
Theorem 4.5.18, The Two-Tangent Angle Theorem: The measure of an angle
formed by two tangents drawn to a circle is one-half the positive difference of the
measures of the intercepted arcs.
In the figures below, both lines are tangent at B and C and
m(  BAC ) 
1
2
x  y
Note: The arc further from A has greater measure. Here, x > y .
Thus, (1/2) | x -- y | =
(1/2) ( x -- y ) since ( x -- y ) > 0 .
D
B
B
2
x
y
y
A
1
 BAC
3
A
C
x
C
= 1 .
m(  2 ) = (1/2) x and m(  3 ) = (1/2) y by Theorem 4.2.16.
m(  2 ) = m(  1 ) + m(  3 ) by Corollary 4.2.3, the Euclidean Exterior Angle Theorem.
m(  1 )
=
m(  2 ) -- m(  3 )
=
(1/2) x -- (1/2) y
= (1/2) ( x -- y ) =
(1/2) | x -- y | .
Corollary 4.5.19, The Two-Tangent-Segments Theorem: Tangent segments
drawn to a circle from the same
A
external point are congruent.
Proof:
Exercise.
[NTS
O
P
PA  PB
]
This is problem #7 in Exercises 4.5 .
B
.
8
Theorem 4.5.20, The Chord Segment Product Theorem: If two chords intersect
within a circle, the product of the lengths of the segments cut in one chord equals
the product of the lengths of the segments cut in the other chord.
That is, (AP)(PB) = (CP)(PD)
Proof:


PA
PD
.

PC
PB
AP
PD
.

CP
PB
[Cross-Multiplying]
(AP)(PB) = (CP)(PD) .
QED
Theorem 4.5.21, The Secant Segment Product Theorem: If two secant segments
are drawn from the same exterior point, then the product of the length of the secant
segment and the length of its external portion is the same for both secants.
That is, (AB)(AC) = (AE)(AD)
Proof:

AB
AD

AE
AC
that
.
[Cross-Multiplying]
(AB)(AC) = (AE)(AD) .
QED
9
Theorem 4.5.22, The Tangent Secant Segment Theorem: If a tangent segment
and a secant segment are drawn to the same circle from the same exterior point,
then the product of the length of the secant segment and the length of its external
portion is equal to the square of the length of the tangent segment.
That is, (AD)(AC) = (AB)2
Proof:

AD
AB

AB
AC
QED
.
[Cross-Multiplying] (AD)(AC) = (AB)(AB) = (AB)2