Download Circles, Arcs and Angles

9/13/2011
Circles, Arcs and Angles
MA 341 – Topics in Geometry
Lecture 09
Star Trek Lemma
• A, B, C are points on a circle centered at O.
• Angle BAC an inscribed angle
• Angular measure of arc BC is the measure of
the central angle BOC, where the angle is
measured on the same side of O as the arc.
B
• BAC subtends the arc BC.
O
A
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Star Trek Lemma
The measure of the inscribed angle is half of
the angular measure of the arc it subtends.
There are several cases to the proof of the
lemma. We will look only at the case where
BAC is an acute angle and the center, O, lies in
the interior of the angle, as in our figure.
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Star Trek Lemma
The measure of the inscribed angle is half of
the angular measure of the arc it subtends.
B
O
C
A
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Star Trek Lemma
Note that OA, OB and OC are all radii
B
O
A
C
Thus, ΔAOB, ΔAOC and ΔBOC are isosceles
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Star Trek Lemma
Extend the segment OA until it meets the circle
at a point D.
B
D
O
A
C
ΔAOB isosceles => BAO = OBA
BOD = 0BA + BAO = 2BAO
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Star Trek Lemma
BOD = 0BA + BAO = 2BAO
B
O
A
D
C
ΔAOC isosceles => CAO = OCA
COD = 0CA + CAO = 2CAO
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Star Trek Lemma
BOD = 0BA + BAO = 2BAO
COD = 0CA + CAO = 2CAO
BOC = BOD + DOC
= 2BAO + 2CAO
= 2BAC
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Why the Star Trek Lemma?
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Circle Theorem
There is exactly one circle through any three
non-collinear points.
A
B
C
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Circle Theorem
A
b
P
B
a
C
Let a be the perpendicular bisector of AC and b
the perpendicular bisector of AB.
Why do they intersect?
Let the intersection be P.
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Circle Theorem
A
X
P
B
Y
C
Construct PA, PB, PC
ΔPBX  ΔPAX – why?  PB  PA
ΔPAY  ΔPCY – why?  PA  PC
 PA  PB  PC
A, B, C equidistant from P. Text shows other
direction
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Circle Theorem
A
B
P
C
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Inscribed Parallelograms
A polygon is inscribed in a circle if all of its
vertices lie on the circle and its interior is
interior to the circle.
Theorem: Opposite angles of an inscribed
quadrilateral are supplementary.
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Inscribed Parallelograms
Theorem: Opposite angles of an inscribed
quadrilateral are supplementary.
A
D
O
B
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Inscribed Parallelograms
A
ABC = ½AOC
ADC = ½AOC’
D
O
B
C
ABC + ADC = ½AOC + = ½AOC’
ABC + ADC = ½(360°) = 180°
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Secants and Angles
Theorem: The angle between two secants
drawn to a circle from an exterior point is
equal to half the difference of the two
subtended arcs.
D
A
O
B
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X
C
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Secants and Angles
D
A
O
X
C
B
AXB = 180 - XAC - XCA
AXB = 180 - XAC – (180 - ACB) = ACB - XAC
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Secants and Angles
AXB = 180 - XAC - XCA
AXB = 180 - XAC – (180 - ACB) = ACB - DAC
AXB = ½AOB - ½ DOA
AXB = ½(
)
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Chords and Angles
Theorem: The angle between two chords is
equal to half the sum of the two subtended
arcs.
AXB = ½(
)
D
A
O
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B
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Circles and Right Angles
Theorem: Given ΔABC the angle at vertex C
is a right angle if and only if AB is a diameter
of a circle.
C
B
A
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Circles and Right Angles
The circumcircle for ΔABC exists. Thus,
ACB is inscribed in the circle and the arc
AB is measured by 2C.
C = 90°  2C = 180°
 AB is a diameter.
C
B
A
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Tangents and Chords
Theorem: The angle between a chord and a
tangent at one of its endpoints is equal to
half the subtended arc.
BAC = ½AOB
O
B
A
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Tangents and Secants
Theorem: The angle between a secant and a
tangent at a point outside the circle is equal
to half the difference of the two subtended
arcs.
B
C
P
BPA = ½(
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A
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