|
|
| Line 1: |
Line 1: |
| In [[geometry]], an '''inscribed angle''' is formed when two [[secant line]]s of a [[circle]] (or, in a [[degenerate case]], when one [[secant line]] and one [[tangent line]] of that circle) intersect on the circle.
| |
|
| |
|
| Typically, it is easiest to think of an inscribed angle as being defined by two [[Chord (geometry)|chords]] of the circle sharing an endpoint.
| |
|
| |
|
| The basic properties of inscribed angles are discussed in Book 3, Propositions 20–22 of [[Euclid's Elements|Euclid's ''Elements'']]. These are the inscribed angle is half the central angle, inscribed angles on the same arc of a chord are equal and the sum of the two distinct inscribed angles of a chord is 180°.
| | Jerrie is what you would call me but While i don't like when [http://www.sharkbayte.com/keyword/regular regular] people use my full nick name. The job I've been taking up for years is any kind of a people manager. Guam is even I've always been living. What I genuinely like doing is fish and I'll be building something else along can. Go to the group website to find out more: http://circuspartypanama.com<br><br>my website ... clash of clans cheats ([http://circuspartypanama.com Recommended Studying]) |
| | |
| ==Property==
| |
| | |
| An inscribed angle is said to intersect an [[Arc (geometry)|arc]] on the circle. The arc is the portion of the circle that is in the interior of the angle. The measure of the [[intercepted arc]] (equal to its [[central angle]]) is exactly twice the measure of the inscribed angle.
| |
| | |
| This single property has a number of consequences within the circle. For example, it allows one to prove that when two chords intersect in a circle, the products of the lengths of their pieces are equal. It also allows one to prove that the opposite angles of a [[cyclic quadrilateral]] are [[Supplementary angles|supplementary]].
| |
| | |
| ==Proof==
| |
| | |
| To understand this proof, it is useful to draw a diagram.
| |
| | |
| ===Inscribed angles where one chord is a diameter===
| |
| [[Image:InscribedAngle 1ChordDiam.svg|right]]
| |
| Let ''O'' be the center of a circle. Choose two points on the circle, and call them ''V'' and ''A''. Draw line ''VO'' and extended past ''O'' so that it intersects the circle at point ''B'' which is [[diametrically opposite]] the point ''V''. Draw an angle whose [[Vertex (geometry)|vertex]] is point ''V'' and whose sides pass through points ''A'' and ''B''.
| |
| | |
| Draw line ''OA''. Angle ''BOA'' is a [[central angle]]; call it ''θ''. Lines ''OV'' and ''OA'' are both [[radius|radii]] of the circle, so they have equal lengths. Therefore triangle ''VOA'' is [[isosceles]], so angle ''BVA'' (the inscribed angle) and angle ''VAO'' are equal; let each of them be denoted as ''ψ''.
| |
| | |
| Angles ''BOA'' and ''AOV'' are [[supplementary angle|supplementary]]. They add up to 180°, since line ''VB'' passing through ''O'' is a straight line. Therefore angle ''AOV'' measures 180° − ''θ''.
| |
| | |
| It is known that the three angles of a [[triangle]] add up to 180°, and the three angles of triangle ''VOA'' are:
| |
| | |
| : 180° − ''θ''
| |
| : ''ψ''
| |
| : ''ψ''.
| |
| | |
| Therefore
| |
| | |
| :<math> 2 \psi + 180^\circ - \theta = 180^\circ. </math> | |
| | |
| Subtract 180° from both sides,
| |
| | |
| :<math> 2 \psi = \theta, \,</math>
| |
| | |
| where ''θ'' is the central angle subtending arc ''AB'' and ''ψ'' is the inscribed angle subtending arc ''AB''.
| |
| | |
| ===Inscribed angles with the center of the circle in their interior===
| |
| [[Image:InscribedAngle CenterCircle.svg|right]]
| |
| Given a circle whose center is point ''O'', choose three points ''V'', ''C'', and ''D'' on the circle. Draw lines ''VC'' and ''VD'': angle ''DVC'' is an inscribed angle. Now draw line ''VO'' and extend it past point ''O'' so that it intersects the circle at point ''E''. Angle ''DVC'' subtends arc ''DC'' on the circle.
| |
| | |
| Suppose this arc includes point ''E'' within it. Point ''E'' is diametrically opposite to point ''V''. Angles ''DVE'' and ''EVC'' are also inscribed angles, but both of these angles have one side which passes through the center of the circle, therefore the theorem from the above Part 1 can be applied to them.
| |
| | |
| Therefore
| |
| | |
| :<math> \angle DVC = \angle DVE + \angle EVC. \, </math>
| |
| | |
| then let
| |
| | |
| :<math> \psi_0 = \angle DVC, </math>
| |
| :<math> \psi_1 = \angle DVE, </math>
| |
| :<math> \psi_2 = \angle EVC, </math>
| |
| | |
| so that
| |
| | |
| :<math> \psi_0 = \psi_1 + \psi_2. \qquad \qquad (1) </math>
| |
| | |
| Draw lines ''OC'' and ''OD''. Angle ''DOC'' is a central angle, but so are angles ''DOE'' and ''EOC'', and
| |
| :<math> \angle DOC = \angle DOE + \angle EOC. </math>
| |
| | |
| Let
| |
| | |
| :<math> \theta_0 = \angle DOC, </math>
| |
| :<math> \theta_1 = \angle DOE, </math>
| |
| :<math> \theta_2 = \angle EOC, </math>
| |
| | |
| so that
| |
| | |
| :<math> \theta_0 = \theta_1 + \theta_2. \qquad \qquad (2) </math>
| |
| | |
| From Part One we know that <math> \theta_1 = 2 \psi_1 </math> and that <math> \theta_2 = 2 \psi_2 </math>. Combining these results with equation (2) yields
| |
| | |
| :<math> \theta_0 = 2 \psi_1 + 2 \psi_2 \,</math>
| |
| | |
| therefore, by equation (1),
| |
| | |
| :<math> \theta_0 = 2 \psi_0. \,</math>
| |
| | |
| ===Inscribed angles with the center of the circle in their exterior===
| |
| [[Image:InscribedAngle CenterCircleExtV2.svg|right]]
| |
| [The previous case can be extended to cover the case where the measure of the inscribed angle is the ''difference'' between two inscribed angles as discussed in the first part of this proof.]
| |
| | |
| Given a circle whose center is point ''O'', choose three points ''V'', ''C'', and ''D'' on the circle. Draw lines ''VC'' and ''VD'': angle ''DVC'' is an inscribed angle. Now draw line ''VO'' and extend it past point ''O'' so that it intersects the circle at point ''E''. Angle ''DVC'' subtends arc ''DC'' on the circle.
| |
| | |
| Suppose this arc does not include point ''E'' within it. Point ''E'' is diametrically opposite to point ''V''. Angles ''DVE'' and ''EVC'' are also inscribed angles, but both of these angles have one side which passes through the center of the circle, therefore the theorem from the above Part 1 can be applied to them.
| |
| | |
| Therefore
| |
| :<math> \angle DVC = \angle EVC - \angle DVE </math>.
| |
| then let
| |
| :<math> \psi_0 = \angle DVC, </math>
| |
| :<math> \psi_1 = \angle DVE, </math>
| |
| :<math> \psi_2 = \angle EVC, </math>
| |
| so that
| |
| :<math> \psi_0 = \psi_2 - \psi_1. \qquad \qquad (3) </math>
| |
| | |
| Draw lines ''OC'' and ''OD''. Angle ''DOC'' is a central angle, but so are angles ''DOE'' and ''EOC'', and
| |
| :<math> \angle DOC = \angle EOC - \angle DOE. </math>
| |
| Let
| |
| :<math> \theta_0 = \angle DOC, </math>
| |
| :<math> \theta_1 = \angle DOE, </math>
| |
| :<math> \theta_2 = \angle EOC, </math>
| |
| so that
| |
| :<math> \theta_0 = \theta_2 - \theta_1. \qquad \qquad (4) </math>
| |
| | |
| From Part One we know that <math> \theta_1 = 2 \psi_1 </math> and that <math> \theta_2 = 2 \psi_2 </math>. Combining these results with equation (4) yields
| |
| :<math> \theta_0 = 2 \psi_2 - 2 \psi_1 </math>
| |
| therefore, by equation (3),
| |
| :<math> \theta_0 = 2 \psi_0. </math>
| |
| | |
| ==Theorem==
| |
| [[Image:Inscribed angle theorem.svg|thumb|right|250px|The inscribed angle ''θ'' is half of the central angle 2''θ'' that subtends the same arc on the circle (magenta). Thus, the angle ''θ'' does not change as its vertex is moved around on the circle (green, blue and gold angles).]]
| |
| | |
| [[File:ArcCapable.gif|thumb|350px|right|Given the two points ''A'' and ''B'', the set of points ''M'' in the plane for which the angle ''AMB'' is equal to ''α'' is an arc of a circle. The measure of the angle ''AOB'', which ''O'' is the center of the circle, is 2''α''.]]
| |
| | |
| The inscribed angle theorem states that an angle ''θ'' inscribed in a circle is half of the central angle 2''θ'' that [[Subtended arc|subtend]]s the same [[Arc (geometry)|arc]] on the circle. Therefore, the angle does not change as its [[Vertex (geometry)|vertex]] is moved to different positions on the circle.
| |
| | |
| The inscribed angle [[theorem]] is used in many proofs of elementary [[Euclidean geometry of the plane]]. A special case of the theorem is [[Thales' theorem]], which states that the angle subtended by a [[diameter]] is always 90°, i.e., a right angle. As a consequence of the theorem, opposite angles of [[cyclic quadrilateral]]s sum to 180°; conversely, any quadrilateral for which this is true can be inscribed in a circle. As another example, the inscribed angle theorem is the basis for several theorems related to the [[power of a point]] with respect to a circle.
| |
| | |
| ===Proof===
| |
| In the simplest case, one leg of the inscribed angle is a diameter of the circle, i.e., passes through the center of the circle. Since that leg is a straight line, the supplement of the central angle equals 180° − 2''θ''. Drawing a [[Circular segment|segment]] from the center of the circle to the other point of [[Line-line intersection|intersection]] of the inscribed angle produces an [[isosceles triangle]], made from two radii of the circle and the second leg of the inscribed angle. Since two of the angles in an isosceles triangle are equal and since the angles in a triangle must add up to 180°, it follows that the inscribed angle equals ''θ'', half of the central angle.
| |
| | |
| This result may be extended to an arbitrarily inscribed angle by drawing a diameter from the vertex of the angle. This converts the general problem into two sub-cases in which a diameter is a leg of each angle. The arbitrary angle equals half of the sum of the two central angles that share the diameter as a leg. Adding the two subangles again yields the result that the inscribed angle is half of the central angle.
| |
| | |
| Note that the central angle for the golden inscribed angle is 360° − 2''θ''. Therefore, the half of it (and thus the measure of the golden inscribed angle) is 180° − ''θ''.
| |
| | |
| The set of all points (locus) for which a line segment can be seen at angle measured ''θ'' contains two arcs (one of each side of the line segment with central angle 2''θ''). In the special case of 90°, there is exactly one circle with center the middle of the line segment.
| |
| | |
| ===Corollaries===
| |
| By a similar argument, the angle between a [[Chord (geometry)|chord]] and the [[tangent]] line at one of its intersection points equals half of the central angle subtended by the chord. See also [[Tangent lines to circles]].
| |
| | |
| ==References==
| |
| * {{Cite book | author = Ogilvy CS | year = 1990 | title = Excursions in Geometry | publisher = Dover | isbn = 0-486-26530-7 | pages = 17–23 | postscript = <!--None-->}}
| |
| * {{cite book | author = Gellert W, Küstner H, Hellwich M, Kästner H | title = The VNR Concise Encyclopedia of Mathematics | publisher = Van Nostrand Reinhold | location = New York | isbn = 0-442-22646-2 | pages = 172 | year = 1977}}
| |
| | |
| ==External links==
| |
| * {{MathWorld |urlname=InscribedAngle |title=Inscribed Angle}}
| |
| * [http://www.mathalino.com/reviewer/plane-geometry/relationship-between-central-angle-and-inscribed-angle Relationship Between Central Angle and Inscribed Angle]
| |
| * [http://www.cut-the-knot.org/pythagoras/Munching/inscribed.shtml Munching on Inscribed Angles] at [[cut-the-knot]]
| |
| * [http://www.mathopenref.com/arccentralangle.html Arc Central Angle] With interactive animation
| |
| * [http://www.mathopenref.com/arcperipheralangle.html Arc Peripheral (inscribed) Angle] With interactive animation
| |
| * [http://www.mathopenref.com/arccentralangletheorem.html Arc Central Angle Theorem] With interactive animation
| |
| | |
| | |
| [[Category:Euclidean plane geometry]]
| |
| [[Category:Angle]]
| |
| [[Category:Circles]]
| |
| [[Category:Articles containing proofs]]
| |