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Central Angle Theorem The measure of an inscribed angle for a ---

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Central Angle Theorem The measure of an inscribed angle for a ---

    Proof of the

    Central Angle Theorem and

    its Corollaries

    April 25, 2006

    Math 408

    Kim Aitken

    Elizabeth Lemons

     The study of Euclidean Geometry is based on the ideas of Euclid of Alexandria and his most famous work, The Elements. Euclid is generally thought of as “the father of

    geometry” and The Elements is considered one of the most successful textbooks in the

    1 The Elements was written over 2,300 years ago and to this day, history of mathematics.

    2no copies are still in existence. The Elements is made up of 13 books, each of which serves as a foundation for all who study geometry today. In Euclid’s third book, he

    discusses circles, segments of circles, and sectors of circles, including the Central Angle

    Theorem and its corollaries.

     In order to fully understand the Central Angle Theorem and its corollaries, we must first state a few definitions. If C(O,r)is a circle, an inscribed angle is an angle of the formPQR, such that P, Q, and R lies on. A central angle is an angle of the

    form, where O is the center of the circle and P and R are on. An inscribed POR

    P angle’s corresponding central angle exists when either “Q and R lie

    3on opposite sides of OP or P and Q lie on opposite sides of OR.”

     Therefore, a central angle only corresponds to an inscribed angle if Q O

    0and only if the inscribed angle has measure less than90. If the

    0inscribed angle is greater than 90, it can be seen from the figure

    R PQRto the right that while inscribed angle intercepts the arc (clockwise from P to PR

    R), central angle intercepts a different arc, the arc PQR (counter-clockwise from POR

    P to R).

     1 "Euclid." Wikipedia, The Free Encyclopedia. 20 Apr 2006, 20 Apr 2006

    <http://en.wikipedia.org/w/index.php?title=Euclid&oldid=49291859>. 2 Allen, G. Donald. Home page. 17 March 2006. 20 April 2006

    <http://www.math.tamu.edu/~don.allen/history/euclid/euclid.html>. 3 Venema, Gerard A., Foundations of Geometry, 239.

     1

    The Central Angle Theorem states that, “the measure of an inscribed angle for a

    4 In order to prove the circle is one half the measure of the corresponding central angle.”Central Angle Theorem, we begin with the hypothesis that is an inscribed angle PQR

    in circle. There are three cases to consider. C(O,r)

    Case 1: O lies on one of the sides of . (See Fig. 1) To begin, assume Q * PQR

    O * R (this is equivalent to the case where the angle makes it R * O * Q). To make the

    notation simpler, we will let a(OQP) and let b(ORP). Since OQ, OR, and

    OP are all radii of the circle and thereby of equal measure, by the Isosceles Triangle Theorem, (OPQ)(OQP)a and (OPR)(ORP)b. By the Angle

    0((QPR)2a;2b180Sum Theorem, . Also by the Angle Sum Theorem,

    0(POR)1802b and by substitution, . Since the (POR)2a;2b2b2a

    inscribed angle (PQR)a and the corresponding central angle(POR)2a, we

    have proven that the measure of the inscribed angle is one half the measure of the corresponding central angle, proving the Central Angle Theorem for this case.

    Case 2: O lies in the interior of PQR. (See Fig. 2) Construct a point S such

    that S lies on and Q * O * S, thereby making S antipodal to Q. By the Protractor

    Postulate, (PQR)(PQS);(RQS) and (POR)(POS);(ROS).

    Again, to make the notation simpler, we will let a(OQP), b(OSP),

    c(OQR)d(OSR), and . Since OQ, OS, OR, and OP are all radii of the circle

    and thereby of equal measure, by the Isosceles Triangle Theorem, (OPQ)(OQP)a(OPS)(OSP)b(ORQ)(OQR)c, , ,

     4 Venema, Gerard A., Foundations of Geometry, 239.

     2

    . By the Angle Sum Theorem, and (ORS)(OSR)d

    00((QPS)2a;2b180((QRS)2c;2d180 and . Also by the Angle Sum

    00(POS)1802b(ROS)1802dTheorem, and . Therefore, by substitution,

     and . Now, (POS)2a;2b2b2a(ROS)2c;2d2d2c

     and (PQR)(PQS);(RQS)a;c(POR)(POS);(ROS)

    . Since the inscribed angle and the corresponding 2a;2c2(a;c)(PQR)a;c

    central angle(POR)2(a;c), we have proven that the measure of the inscribed angle is one half the measure of the corresponding central angle, proving the Central

    Angle Theorem for this case.

     Case 3: O is neither on PQRor in the interior. (See Fig. 3) Construct a

    point S such that S lies on and Q * O * S, thereby making S antipodal to Q. By the

    Protractor Postulate, (PQR)(PQS)(RQS) and

    (POR)(POS)(ROS). Again, to make the notation simpler, we will let a(OQP), b(OSP), c(OQR), and d(OSR). Since OQ, OS, OR,

    and OP are all radii of the circle and thereby of equal measure, by the Isosceles Triangle

    Theorem, (OPQ)(OQP)a, (OPS)(OSP)b,

    (ORQ)(OQR)c, and (ORS)(OSR)d. By the Angle Sum

    00((QPS)2a;2b180((QRS)2c;2d180Theorem, and . Also by the

    00(POS)1802b(ROS)1802dAngle Sum Theorem, and . Therefore, by

    (POS)2a;2b2b2a(ROS)2c;2d2d2csubstitution, and . Now,

    (PQR)(PQS)(RQS)ac(POR)(POS)(ROS) and

    2a2c2(ac)(PQR)ac. Since the inscribed angle and the corresponding

     3

    , we have proven that the measure of the inscribed central angle(POR)2(ac)

    angle is one half the measure of the corresponding central angle, proving the Central Angle Theorem for the final case. By proving all three of these cases, we have proved the Central Angle Theorem.

     One of the important corollaries that follows from the Central Angle Theorem is Corollary 10.4.2. It states, “if the vertices of triangle lie on a circle and is a ABABC

    5diameter of that circle, then <ACB is a right angle.” This corollary is a special case of

    the Central Angle Theorem and the similarities can be seen in the proof of the theorem. To begin, we must state our hypotheses that A, B, and C lie on circle C(O,r) and A *

    O * B. Again, to make the notation simpler, let a(OAC) and let b(OBC).

    Since OA, OB, and OC are all radii of the circle and thereby of equal measure, by the Isosceles Triangle Theorem, (OCA)(OAC)a and

    (OCB)(OBC)b. By the Protractor Postulate,

    (ACB)(OCA);(OCB)a;b. Then by the Angle Sum Theorem,

    000((ABC)2a;2b180(ACB)90a;b90, which implies that . Therefore,

    and is a right angle. This completes the proof of Corollary 10.4.2. ACB

     While most people consider geometry to be mainly about lines and triangles, Euclid worked with both the definitions and theorems on circles, lines and triangles from

    6the start, giving them all equal treatment. Within the study of circles is the Central

    Angle Theorem and its corollaries. The study of circles has also continued to grow with more modern mathematicians and technology adding to the field. While The Elements is

     5 Venema, Gerard A., Foundations of Geometry, 238. 6 Venema, Gerard A., Foundations of Geometry, 225.

     4

    highly respected, “several of the propositions are regarded as not entirely satisfactory today. The proofs of some seem to be lacking in clarity, if not rigor; some rely on unproved assumptions; and others suffer from Euclid’s tendency to avoid multiple cases in his proofs. In any event, modern textbooks tend to present the material on circles

    7 The Central Angle Theorem, having multiple cases, may have differently from Euclid.”

    been one of the theorems that needed expanding by modern mathematicians since the discovery of The Elements. Despite the fact that we are uncertain about all of the propositions in The Elements, Euclid’s ideas still established the foundation for all of Euclidean geometry, including these important theorems on circles.

     7 Lancon, Donald Jr., An Introduction to the Works of Euclid with an Emphasis on The Elements. 04 Oct. 2003. 20 April 2006. <http://www.obkb.com/dcljr/euclid.html>

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Fig. 1:

    , 180 = 2a + 2b P From PQR

     <PQR = a and

     a b <POR = 180-2b = 2a + 2b 2b = 2a

    0 For Corollary 10.4.2, <QPR= a + b= 90

     Q a b R

     O

Fig. 2:

    PQS, 180 = 2a + 2b From P

    <PQS = a

    <POS = 180 2b = 2a + 2b 2b = 2a a b

     From RQS, 180 = 2c + 2d

     Q a O b S <RQS = c

     c d <ROS = 180 2d = 2c + 2d 2d = 2c

     So <PQR = <PQS + <RQS = a + c and

     c d <POR = <POS + <ROS = 2a + 2c = 2(a+c)

     R

Fig. 3:

     R 180 = 2c + 2d

     P c d <RQS = c

     a b <ROS = 180 2d = 2c + 2d 2d = 2c

     180 = 2a + 2b

     Q c a d b S <PQS = a

     O <POS = 180 2b = 2a + 2b 2b = 2a

     So <PQR = <PQS - <RQS = a c and

     <POR = <POS - <ROS = 2a2c= 2(a-c)

     6

    Bibliography

     Allen, G. Donald. Home page. 17 March 2006. 20 April 2006.

    <http://www.math.tamu.edu/~don.allen/history/euclid/euclid.html>

Lancon, Donald Jr., An Introduction to the Works of Euclid with an Emphasis on The

    Elements. 04 Oct. 2003. 20 April 2006. <http://www.obkb.com/dcljr/euclid.html>

     Venema, Gerard A. Foundations of Geometry. New Jersey: Pearson Prentice Hall,

    2006.

Wikipedia, The Free Encyclopedia, “Euclid. 20 Apr 2006, 20 Apr 2006.

    <http://en.wikipedia.org/w/index.php?title=Euclid&oldid=49291859>

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