Download Circles in Euclidean Geometry, Part I

Section 4.5: Circles in Euclidean Geometry (Part 1: Through Corollary 4.5.13)
Definition: Given a point P and a real number r > 0 , the circle C( P, r ) is the set
of points with a common distance r from point P, i.e., the set of points Q in the
plane such that PQ = r .
Point P is the center and number r is the radius of the circle C( P, r ) .
Theorem 4.5.1, The "Three Points Circle" Theorem:
In the Euclidean Plane, three distinct non-collinear points determine a unique circle.
Proof: Let A, B, and C be three non-collinear points. Let l and m be the
perpendicular bisectors of AB and BC , resp.,
intersecting these segments at midpoints M and N,
resp. .
Suppose l is parallel to m. By Theorem 3.4.8, if a
line is perpendicular to one of two parallel lines,
then it is perpendicular to the other. Therefore,
suuur
suuur
since AB ⊥ l , AB ⊥ m. Let D be the point where
AB intersects m . D is not equal to N or else A, B,
and C would be collinear. Thus, ∆ BDN has angle
sum greater than 180°, which is a impossible in
Euclidean Geometry. Therefore, l is not parallel to
m.
m
l
A
C
N
M
B
D
Therefore, l and m intersect, say a point P. Since
P is on the perpendicular bisectors of segments,
PA = PB and PB = PC by Theorem 3.2.8.
Let r = PA = PB = PC . Then, all three points,
A, B, and C are on the circle C( P, r ). Any other
possible circle passing through A, B, and C must
have its center on both perpendicular bisectors and
so its center must be the point P of intersection of
these perpendicular bisectors, so there is no other
such circle than C( P, r ) passing through A, B, and
C.
QED
P
l
m
C
A
N
M
B
Note: This proof provides a method for constructing the center of the circle.
2
Facts from Theorems (Proofs left as exercises):
1) The "Longest Chord" Theorem:
The Diameter of a circle is a chord with length = 2r, which is longer than any nondiameter chord of the circle (Theorem 4.5.3).
2) In a circle C( O, r ) with a diameter
AB
and a chord
PQ
,
a) The "Chord Bisector" Theorem:
AB ⊥ PQ
if, and only if,
AB
bisects
PQ
(i.e., they intersect at the midpoint of
PQ
).
b) The "Chord Perpendicular Bisector" Theorem:
The perpendicular bisector of any chord
PQ
will contain the center O .
(The statement (a) combines Theorem 4.5.4 and Theorem 4.5.5.
Statement (b) is Theorem 4.5.6.)
Here is the proof of that part of (a) which says “If
AB ⊥ PQ
, then
AB
bisects
PQ .”
Proof:
Suppose
AB
is a diameter of C( O, OA) and
Note: OP and OQ are both radii.
GPO ≅
PQ
is a chord such that
A
Q
r
GQO by HL .
O
∴ PG = GQ by CPCF .
G
r
B
∴ G is the midpoint of PQ .
QED
AB ⊥ PQ .
P
Note how the word “radius” is used in two ways, both as the number which is the
common distance from the center and as a line segment from the center to a point
on the circle.
3
Theorem 4.5.7, The "Tangents are Perpendicular" Theorem
Theorem:
If a line is tangent to a circle, then it is perpendicular to the radius drawn to the
point of tangency.
Proof: Assume that line t is tangent to
circle C( O, r ) with point of tangency at point
P. Suppose that t is not perpendicular to OP .
Draw a perpendicular segment from O to point
Q on line t . Then, locate point R on line t
with P-Q-R
R and with PQ = QR .
∆OQP ≅ ∆OQR
R
Q
O
r
P
by SAS .
∴ OR = OP = r , by C P C F .
∴ Point R is on the circle and t is not a
tangent, which is a contradiction.
∴ , line t
⊥ OP .
t
QED
Definitions:
Central Angle, Chord corresponding to a central angle, Minor arc, Major arc,
arc
Inscribed angle, Intercepted arc, Tangent segment.
The measure of AB is: 1) 180
180° if
AB
is a diameter of the circle.
2) m( ∠ AOB ) if the arc is a minor arc .
3) 360
360°– m( ∠ AOB ) if the arc is a major arc .
4
Facts:
3) Two minor arcs on the same circle are congruent arcs if, and only if, their
corresponding chords are congruent line segments (Theorems 4.5.8 & 4.5.9) .
4) Arcs (both “major + minor” and “minor + minor”) and their measures add up
appropriately (Theorem 4.5.10).
Def’ns: “An angle inscribed in Arc XYZ” and “An angle intercepts Arc XYZ”
∠ APB is inscribed in the Arc APB .
∠ APB is intercepts Arc ARB .
( We will also say that
"
∠ APB
is inscribed in circle C( O , OR )" .)
An angle intercepts an arc on a circle when:
1) the arc endpoints are on the rays of the angle, and
2) the interior of the arc is contained in the interior of the angle.
There are several ways that angles intercept arcs:
In all of these examples,
∠ APB
intercepts arc ARB and also the marked arcs .
In the third example, segment PB is called a Tangent Segment.. In the fourth
example, segments PB and PA are both tangent segments.
5
Theorem 4.5.11, The "Inscribed
Inscribed Angle
Angle" Theorem : The measure of an angle
inscribed in an arc is one-half
half the measure of its intercepted arc.
[Translated: The angle measure of an angle inscribed in an arc is the ½ times the arc
measure of the arc which it intercepts (i.e., the arc which is in the interior of that
angle). This measure is ½ times the degree measure of the central angle which
intercepts that arc.]
APB in circle C( O, OP ) with A and B on
Proof: Let ∠ APB be inscribed in arc A
this circle. There are three cases to consider.
Case 1:
PA
or
PB
is a diameter of the circle and the center O is on that segment.
∴ m(
OP = OA = radius .
∴ m(
∠ 4
) = m(
∴ m(
∠ 1
) = ½ m(
∠ 1
) + m(
∠ 4
∠1
∠ 2
) = m(
∠ 2
) = 2 m(
)
∠1
)
) = ½ m( arc AB ).
QED for CASE 1.
Case 2: The center O is in
the interior of
∠ APB
Case 3: The center O is in
.
the exterior of
∠ APB
.
x°
C
y°
[ NTS m( ∠ APB ) = ½ ( x° + y° ) ]
m( arc ACB ) = ( x° + y°° )
m( ∠ APB ) = m( ∠ APC ) + m( ∠ CPB )
= ½ x° + ½ y°
(from Case 1)
m( ∠ APB ) = ½ ( x° + y° )
[ NTS m( ∠ APB ) = ½ y° ]
m( arc ABC ) = ( y°
y + x° )
m( ∠ APB ) = m( ∠ APC ) – m( ∠ BPC )
= ½ ( y° + x° ) – ½ x°
m( ∠ APB ) = ½ y°
y
(from Case 1)
QED
6
Corollary 4.5.12,, The "Angle Inscribed in a Semicircle" Theorem:
Theorem
An angle which is inscribed in a semicircle is a right angle.
Proof: Draw the ray PO.
The m( ∠ APB ) = m( ∠ APC ) + m( ∠ CPB )
= ½ m( arc AXC ) + ½ m( arc CYB )
= ½ ( m(( arc AXC ) + m( arc CYB ) )
= ½( 180°° ) = 90° .
∴ m( ∠ APB ) = ½( 180°° ) = 90° QED
Theorem (NIB) 4.4,, The "Perpendiculars are Tangent" Theorem:
Theorem
The Converse of Theorem
heorem 4.5.7 (Exercises
Exercises 4.5, Problem 9) :
If a line intersects a circle at a point and the line is perpendicular to the radius
(or diameter) drawn to that point, then the line is
tangent to the circle at that point.
l
Proof: Suppose that line l intersects circle
C( O, r ) at point A, and
nd suppose also that l is
perpendicular to the radius OA drawn to the
point of intersection,
A. Extend the segment AO
uuur
along
the ray AO to the point B where the ray
uuur
AO intersects circle C( O, r ) . Then, segment
AB is the diameter of the circle drawn to point
A and line l is perpendicular to this diameter
also.
A
r
C
O
B
Suppose, by way of contradiction, that line l
is not tangent to circle C( O, r ) at point A. Then, there is a second point C
where line l intersects circle C( O, r ) . Then, ∠ ACB is inscribed in a semicircle,
and so, by Corollary 4.5.12, ∠ ACB is a right angle. Also, ∠ BAC is a right angle
since line l is perpendicular to the diameter AB . Thus, ∆ ACB has two right angles,
one at A and another at C, and the angle at B has positive measure, so the angle
sum of ∆ ACB is greater than 180
180°. This contradicts that fact that the angle sum of
any triangle in Euclidean Geometry is equal to 180
180°.
Therefore, line l is tangent to circle C( O, r ) at point A. QED
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Corollary 4.5.13, The "Angles inscribed in Common Arc" Theorem:
Theorem
Angles inscribed in the same or congruent arcs are congruent .
(This will mean that two inscribed angles in a circle which intercept the same arc
are congruent.)
Proof: Suppose ∠ APB and ∠ AQB are both inscribed in the same arc, arc APB, on
circle C( O, OA ) and so they both intercept the same arc, arc ARB.
By the Inscribed Angle Theorem,
Theorem 4.5.11,
m(
∠ APB
) = ½ m( arc ARB ) and
m(
∠ AQB
) = ½ m( arc ARB ) .
Therefore, m( ∠ APB ) = m( ∠ AQB ) by
substitution. ∴ ∠ APB ≅ ∠ AQB .
QED
Also, note that, if two angles inscribed in the same circle intercept the same arc,
then the two angles are inscribed in the same arc of the circle and
by Corollary 4.5.13, the two inscribed angles are congruent.
Therefore, these observations essentially form the proof of
Theorem (NIB) 4.5,, The "Intercept Same Arc" Theorem
Theorem:
If two angles inscribed in the same circle intercept the same arc,
then the two angles are congruent.
Proof: See the observations in the previous paragraph.